Hi,
I wonder if anyone can help me with the following please?

Two variables x and y are connected by the law y = a^x. The graph of log 4 y against x is a straight line passing through the origin and the point (6,3). Find the value of 'a'.

First I converted the y = a^x into log a y = x, but got nowhere substituting (6, 3) into the equation.
Next I tried to take logs of both sides:
3 = a^6 log 3 = 6 log a log a = log 3/6 but this gave me the answer 1.2 (the answer is 2)

Can anyone direct me to which method to use to solve this?

Thanks

RATHER A TRICKY QUESTION!

(You should have typed log 4 y as log y (base 4) to avoid confusion. I thought you meant log (4y) and had to try both alternatives)

Suppose coordinate system is (Y,x)

Then the equation of the line passing through the two given points is Y = x/2

Let Y = Log (base 4) y

Equation of line is Y = x/2

y = a^x

(Base for Log is 4 all through)

log y = x log a i.e. Y = x log a (Y = log y)

Y = x log a shoud be identical to Y = x/2

Hence log a = 1/2

Base is 4

a = 4^1/2 = 2

Let me know if any step not clear.

the answer is wrong.
y = a ^ x.
substitute.
0 = a ^ 0. wrong.
0 <> a ^ 0 = 1.
any real to zero power equals one!
they must have typoed someplace.
3 = a ^ 6?

coordinate system is (x,y).
maybe they again got order wrong. And they forgot about a ^0 = 1.

Quote:

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Originally Posted by xxzzy

Hi,
I wonder if anyone can help me with the following please?

Two variables x and y are connected by the law y = a^x. The graph of log 4 y against x is a straight line passing through the origin and the point (6,3). Find the value of 'a'.

First I converted the y = a^x into log a y = x, but got nowhere substituting (6, 3) into the equation.
Next I tried to take logs of both sides:
3 = a^6 log 3 = 6 log a log a = log 3/6 but this gave me the answer 1.2 (the answer is 2)

Can anyone direct me to which method to use to solve this?

Thanks

---

(See my Solution given earlier)

The important point is that this is a case of coordinate transformation.

In many real-life problems, variables (eg x, y) do not have a desired functional relation while transformation of a variable (in this case Log y (base 4)) can result in a desirable functional relation (eg Straight line, Linear, etc.)

I am now very confused. Is Jiten 55's response correct or am I to understand from eawoodall that there is a typo in the question?
And what does the statement "You must spread some Reputation around before giving it to jiten55 again" as a rating mean?

Main thing for you is to make sure you understand every step of a solution.

NOTE the fact that the graph is not of y against x BUT of Log y (base 4) against x. The question does NOT say that (0, 0) lies on y = a^x.

If you understand every step of an answer, and every step makes Mathematical sense to you, then you can trust the answer!

As I wrote, if you have difficulty in understanding any step, do ask for assistance and I will try to clarify.

In Mathematics, if every step is correct then the conclusion has got to be correct!

This is unlike subjects like history where it is often a matter of opinions.

As regards some "comment" on "ratings", again it is entirely up to you. If YOU like
a response, you have every right to give it a good rating - because someone took time to help you.

BUT NEVER REST until you understand each step of a Solution. Otherwise you will never be good at Mathematics!

ANOTHER Solution (Perhaps easier)

ALL logs are taken with base 4

y = a^x (Equation 1)

Hence log y = x log a (Equation 2)

(0, 0) lies on the graph of Log y against x

Hence, when x =0, log y = 0 (This condition is automatically satisfied by equation 2)

(6, 3) lies on the graph of Log y against x.

Hence, when x = 6, Log y = 3

Substitute in Equation 2:

3 = 6 log a

log a = 1/2

(Base is 4)

a = 4^(1/2) = 2

Hence a = 2

Thanks Jiten 55 that explanation was totally clear.
Sorry I couldn't give you a high rating but that statement' You must spread etc... ' was there.
Thank you again. I hope I can come back to you with a question in the future !

the explanation of the question was wrong.

I answered what they asked.

I am glad you saw through what they asked to see what they wanted.

but I was not wrong. You are wrong for presuming so jiten55.

what I did was take your equation 1, and make it into this:

log (base a) y = x (equation 3)
which has to be mathematically true if equation 1 is true, always!
they said it went through origin, which it cannot because the origin is (0,0).
and for no base a is 1=0.

secondly they said it was a straight line. Log are not straight lines.

the related equation you showed is a straight line.
they did not ask the right type of question, because they do not understand that they are dealing with more than one line. A log line, and a related line that is straight.

glad you were able to fix their problem with their understanding what the question should have been.

but of course we are not suppose to do people homework problems for them, just find a way to show them how to do it themselves. And hopefully your example will show them how to do similar ones. Personally I think the problem needs to be reworded.

Quote:

---

Originally Posted by jiten55

ANOTHER Solution (Perhaps easier)

ALL logs are taken with base 4

y = a^x (Equation 1)

Hence log y = x log a (Equation 2)

(0, 0) lies on the graph of Log y against x

Hence, when x =0, log y = 0 (This condition is automatically satisfied by equation 2)

(6, 3) lies on the graph of Log y against x.

Hence, when x = 6, Log y = 3

Substitute in Equation 2:

3 = 6 log a

log a = 1/2

(Base is 4)

a = 4^(1/2) = 2

Hence a = 2

---

XXZZY, You are very welcome.

Your questions have been very interesting.

I love questions that are NOT very easy to solve!

Dear EEWOODALL:

This was a rather tricky question. I almost made the same error as you.

What you missed was that the graph is between Log y and x, not between y and x.

This question shows how important it is to read a question carefully before attempting an answer.

One of my professors used to emphasise: UNDERSTAND IT BEFORE YOU SOLVE IT.

okay, as I said, the question as I understand it is poorly worded, but that is not the fault of those asking. I think it is poor understanding of what the teacher or other who phrased it.

all that matters in the end is the person asking question can use either example, mine or yours to understand how to get the answer, and if they understand yours better, that is cool. I was trying to be more exact, and perhaps a little picky.
so if you xxzzy understand that we both are trying to help, that is all/

---------
note (for doing any log problem):
I always solve log problems the same way.
start with what you know, and change from log to exponent form, and back.
2^3=8.
write down the base, and switch everything else.
log (base 2) 8 = 3.
use the rules of logs to change the way it is written.
log (base 2) 2^3 = 3.
3 log (base 2) 2 = 3.
3/3 log (base 2) 2 = 3.
log (base 2) 2 = 1.
switch back to exponent form,
by writing down the base of the exponent and switching eveything else.
2 ^1 = 2.
done.
------------

I think the problem should be stated as :

n ^ x = a ^ y. (my equation 2)
given that
log (base n) (a^y) = x. (my equation 1)

then you solve easily for a given that you know other values.
base is given as 4.
(x,y) is given as both (0,0), and (3,6).

so
4 ^3 = a ^ 6. (note: and x=0 and y =0 also has to work).
or restated
log (base 4) a^6 = 3.
6 log (base 4) a = 3.
6/3 log (base 4) a =1.
2 log (base 4) a = 1.
log (base 4) (a^2) = 1.
4 ^ 1 = a ^ 2.
a^2 = 4.
a ^2 = 4 = 2^2.
a = 2.

those are the steps I would use.
do not say:
log (base 4) y relates to some related relationship of x. for base 4 where x and y can have a table of values of (0,0) or (x,y). It is sloppy in my humble opinion.
but I guess that is just the methodology by which this concept is usually taught.

at no time was log y = x.
instead realize that log (base 4) a^y = x.
I think it is sloppy math to write it as the first way because it is a related but not the actual equation.

but I understand why people do so. And your explanation is okay. But mine is better.
I hope seeing it several ways helps the person who asked.

-------------------------
jitten55:
as an aside: (please see my explanation above and see if you disagree with any of it)

I just think your explanation and solution could use a little cleaning up.

what you missed is (x,y) is always what is inside the parathesis, never (x,log y), always the values inside (d,f) are (x,y).
graphs are always between (x,y).
this is a straight line graph, they said so in the question.
a log graph is not a straight line.
logorithms are not straight.
you did not have the base in you equation. Remember it is base 4. if you do not include base it is base 10 assumed.

--------
jitten you gave:
y=a^x. then you think plug in the numbers given
when y =0 x=0, true.
when y = 3 x=6? False! You think 6 = 2^3?
again there are two bases, n and a. see my solution. Because it is exactly right.
if I did not mess up. Hopefully I did not. But I know the way I solved it is valid. Unlike your sloppy method. Please try to be exact this is math.

Dear eewoodall:

Let us agree to disagree!

No typo. I was pointing out another way to think of the question.
And solve the question. (one that is not as bad in my humble opinion, and correctly does all the math). If you can understand what jitten meant to say, and how to solve the problem that is all that matters. I was pointing out several parts that I thought were not stated as well as I think they should be.

Quote:

---

Originally Posted by xxzzy

I am now very confused. Is Jiten 55's response correct or am I to understand from eawoodall that there is a typo in the question?
And what does the statement "You must spread some Reputation around before giving it to jiten55 again" as a rating mean?

---

math in our regular system of numbers is set and closed by convention.
there is no way to disagree.

given the starting values:
you said y = a^x.
did you say? 3=a^6. or 6= a^3?
either way if a is 2 or more, you are wrong!
the math statements are false, get over the fact that you are doing well to explain the problem, but you are doing bad sloppy math to do so!

instead consider that if you write or think of the problem as multiple base equation, you can easily solve it, and always do right math.
n^x=a^y. then it is just a question of prime factoring the components of the resultants of the exponentation.
i.e.. n=(certain prime factors multiplied together), and so on with x,a,and y.
then it is easy to see what parts have to be missing.
and the rules of math can be maintained.
at least I know how to spell your login name jitten55.

Quote:

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Originally Posted by jiten55

Dear EEWOODALL:

Let us AGREE to DISAGREE!

---

man have I been stupid. I was wrong. I understand now why I was wrong. And how I was wrong. I hope I have it all correct this time. Sorry I was not understanding every step and part of it before. Here is my hopefully best and last attempt at it.
I apologize to the asker and jitten55, thank you for being waiting for me to 'get it'.

so
the graph of "log (base 4) y against x" means:
log (base 4) y is related to x by the rule:
log y = x log a. (where the base is 4).
and when log (base 4) y = 0 x = 0.
and when log (base 4) y = 3, x = 6.
when is log (base 4) y = 0 ?
change to exponent form:
4^0 = y.
ding ding! 4^0= 1.
when log y =0, y = 1.
log y = x log (base 4) a.
0 = 0 log (base 4) a.

note to self:
2^3=8.
log (base 2) 8 = 3.
p^0=1.
a^x=y. implies:
log y = log a^x.
log 8 = log 2^3.
log 8 = 3 log 2.

given:
y=a^x.
(takes logs of both sides)
log y = log a^x.
log y = x log a. (plug in 3 for log y, and 6 for x.)
3 = 6 log a.
3/6 = log (base 4) a.
log (base 4) a = (1/2).
note: (change to exponent form. By writing down exponent and switching everything else.)
4^(1/2) = a.
2=a.

that part after the last 'given' was what I should have written days ago, again sorry it took me so long to understand. Thank you for helping me learn.

what confused me was what confused the questioner, (x,log y) instead of (x,y).
--------------------------------------
xxzzy said :

Two variables x and y are connected by the law y = a^x. The graph of log 4 y against x is a straight line passing through the origin and the point (6,3). Find the value of 'a'.

First I converted the y = a^x into log a y = x, but got nowhere substituting (6, 3) into the equation.
Next I tried to take logs of both sides:
3 = a^6 log 3 = 6 log a log a = log 3/6 but this gave me the answer 1.2 (the answer is 2)

Can anyone direct me to which method to use to solve this?

Thanks

Quote:

---

Originally Posted by jiten55

ANOTHER Solution (Perhaps easier)

ALL logs are taken with base 4

y = a^x (Equation 1)

Hence log y = x log a (Equation 2)

(0, 0) lies on the graph of Log y against x

Hence, when x =0, log y = 0 (This condition is automatically satisfied by equation 2)

(6, 3) lies on the graph of Log y against x.

Hence, when x = 6, Log y = 3

Substitute in Equation 2:

3 = 6 log a

log a = 1/2

(Base is 4)

a = 4^(1/2) = 2

Hence a = 2

---

Dear EAWOODALL:

There is nothing wrong with "being wrong"!

We all have been wrong one time or another.

I am often wrong.

That's the only way we learn.

BTW, Benjamin Franklin, in his autobiography says (I don't recall the exact words):

"He gave me excellent advice when I was young. He told me I talked as if I was always absolutely right. He advised me to express my opinions in a tentative manner, like this: 'I may be wrong, and I am frequently wrong, but this is what I think... ' "

AND Dale Carnegie says ("How to win friends and influence people") :

"Never tell anyone they are wrong".


Best wishes

Jiten

Furthermore, for the Mathematically more mature, I think my first solution was the more elegant.

Dear EAWOODALL:

You may wish to rethink your following steps:

log y = x log a. (where the base is 4).
and when log (base 4) y = 0 x = 0.
and when log (base 4) y = 3, x = 6.
when is log (base 4) y = 0 ?
change to exponent form:
4^0 = y.
ding ding! 4^0= 1.
when log y =0, y = 1.

INSTEAD I WOULD WRITE:

log y = x log a. (where the base is 4).

When x = 0, log y is zero.

i.e. log y = 0 when x = 0

(Regardless of value of a)

The point being that (0, 0) gives us no clue to the value of a, because (0,0)
lies on the graph whatever be the value of a.

the related equation is different than the equation of y =a^x.
the related equation is a straight line.
since we solved for 'a', we know it is 2.
y = 2^x.
lets plug in the independent variable numbers that we were given;
i.e. valid vallues for x.
y = 2^0.
y=1.
see below if you want to prove it in 'log y' form.

when is log (base 4) y = 0 ?
change to exponent form:
4^0 = y.
ding ding! 4^0= 1.
when 'log y' =0, y = 1.

y = 2^6.
y = 64.
or in 'log y = 3' form.
log (base 4) y = 3.
4^3 = y = 64.

so we have a related graph that is a straight line going through
(0,1) and (6,64).

y=mx+b. for all straight lines. M = (y2-y1)/(x2-x1).
y = (64-1)/(6-0) * x + 1.
y = (63/6) * x + 1.
this is the straight line the equation spoke about.
it is related and formed by the logorithm information given.

y=2^x. plug in numbers and prove it is not a straight line.
x = 0, what does y =? 1.
x = 1, what does y =? 2.
x = 2 what does y =? 4.
x = 3 what does y =? 8.
you can see that these points are not all on the same line by:
y=mx+b.
y = mx +1.
y = (2-1)/(1-0) * x + 1 = 1/1 *x +1 = x +1.
y = (8-4)/(3-2) * x + 1 = (4/1) *x +1 = 4x +1 .
so y = a^x is not and can never be a straight line, which was my argument several posts ago. We only determined the straight line that is related to y = 2^x.
if the teacher had asked for that equation then we could have done all this that I have done in addition to what was asked. They only wanted to know what 'a' was.

--------------------
or perhaps
the teacher wanted a graph of a straight line that is defined by:
(x,'log y') , in which case, the graph would be:
defined by (0,0) (6,3)
so the vertical axis would be labeled "log y". Or to avoid confusion call it z.
the horizontal axis would be labeled "x"
the points on this graph to define the line would be (x,z).
z = (3-0)/(6-0) * x + 0.
z = 3/6 * x + 0.
z = x/2.

a standard graph is labeled on the vertical axis y.
a standard graph is labeled on the horizontal axis x.
perhaps that is also part of the way the problem was set up. That the teacher did not want standard labeling or standard conventions (x,y). It is confusing.