Ask Me Help Desk - Is there a solution to 9^m + 3^(3-2m) = 28?

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Is there a solution to 9^m + 3^(3-2m) = 28?

I don’t think there is. It has been given as a question but I always understood that unlike exponents couldn’t be added.
I can re-write it as 3^2m + 3^(3-2m) = 28 but I’m not sure how that would help me.
Any help would be most appreciated.
Cheers,
Adrian

Actually, it is solvable. Start by noting that:

$3^{(3-2m)} =\frac {3^3}{3^{2m}}=\frac {27}{9^m}$

Then multiply the equation by $9^m$ . This gives an equation with terms having factors of $9^m$ and $9^{2m}$ . At this point you can substitute a new variable for $9^{m}$ ; for example let $x=9^{m}$ . Now you have a quadratic equation in x, which you can factor to solve for x. Finally convert back using $\log 9^m = log x$ , so:

$m = \frac {\log x}{\log 9}$

I suggest you use log base 3 for that. Good luck, and post back with your solutions (there are two).

Wow! Okay, so I got to (9^m-1)(9^m-27)=0 and so the solutions are m = 0 or m= 3/2. Correct? When I substituted the values, they both worked in the original equation. Thanks so much.

Quote:

Originally Posted by AdrianCavinder

the solutions are m = 0 or m= 3/2. Correct?

Yes - correct!

I just noticed that the question said = 28. Not 27.

Quote:

Originally Posted by CravenMorhead

I just noticed that the question said = 28. Not 27

Right - that's why it factors the way that Adrian showed. If the problem had been =27 the solution would have been a bit messier.

Been too long since I did this kind of math. Oy. Just looked off.

Okay, I’m afraid I’ve hit upon another stumbling block.
The original question is: log10(8x) - log10(1+sqrtx) = 2
I got it to 10^2 = 8x/(1 +sqrtx)
Now I get confused.
If I multiply both sides by (1 + sqrtx) I get
100 + 100sqrtx = 8x
Can you let me know if I’m on the right track?
Thanks.

Yes, you're on the right track. Substitute w = sqrt(x) and see where it takes you. Warning: it's a bit of a mess.

Oh gosh, that’s a step I forgot about. So substituting I get 100 + 100w = 8w^2
Using the quadratic formula, I got [100 +/- sqrt(100^2 + 320)]/16. So the result for w was 13.43 or -0.93. Substituting back and squaring, I finally arrived at x = 180.37 or x = 0.87. How did I do?

Quote:

Originally Posted by AdrianCavinder

I finally arrived at x = 180.37 or x = 0.87. How did I do?

Pretty good, except... try substituting those values back into the original equation and see if they work. One does, but the other does not.

Right. I got w = -0.93 & 13.43 and when substituted back as sqrtx, the former is not possible, so there’s only one answer which is (13.43)^2 or x = 180.37. Really appreciate your input. Thanks a lot.

Okay. Hi and thanks again for the help.
Needless to say I have one more question that needs a bit of input.
The population of New York and LA aregrowing at 1% and 1.4% a year. Starting from 8 million (NY) and 6 million(LA), when will they be equal?
My reasoning is that this works [FONT=CMR10]similar to compound interest, so
6,000,000 x (1.014^n) = 8,000,000 x (1.01^n)
This can be simplified to
0.75 x 1.014^n = 1.01^n
Now I’m stuck again. I’d appreciate a pointer or two but please don’t solve it for me. How can I deal with unlike bases with the same indices?
Cheers. Adrian[/FONT]

Take the log of both sides, and simplify to get n by itself.

Okay, well, honestly that’s confusing. Or should I say asks more questions than it solves, ha!
1). Why would I use logarithms in a situation like this?
2). Can it be solved without the use of logarithms?
3). My limited understanding of logarithms is using the basic formula of a^x = y is the same as loga(y) = x. How am I then allowed to ‘log’ both sides? Surely, if I substitute this formula into my original equation, I would get x = 1.01^n? And that seems to complicate matters.
Sorry to sound like a numbskull but I want to understand the link before I just plough ahead and arrive at an answer without knowing why.
Cheers, Adrian

Quote:

Originally Posted by AdrianCavinder

Okay, well, honestly that's confusing. Or should I say asks more questions than it solves, ha!
1). Why would I use logarithms in a situation like this?

Logarithms are a usefull tool for working problems involving exponents.

Quote:

Originally Posted by AdrianCavinder

Can it be solved without the use of logarithms?

The only other technique i can think of is to use a numerical technique - essentuially making a series of guesses as to the answer and zeroing in until the error is "acceptable," although you may not get an exact solution this way.

Quote:

Originally Posted by AdrianCavinder

My limited understanding of logarithms is using the basic formula of a^x = y is the same as loga(y) = x. How am I then allowed to 'log' both sides?

You can start with:

$\log (0.75(1.014^n)) = \log (1.01^n)$

Recall that $\log (ab) = \log a + \log b$ and $\log a^b = b \log a$ ; hence the equation can be rewritten as:

$\log (0.75) + n \log (1.014)= n \log (1.01)$

Can you finish it up from here?

Yep, that’s a real big help. Thanks a bundle.
Working out the logs, then subtracting the ’n’ values, then dividing by log(0.75), I got 72.75 years which checks when I replace the values in the original equation.
Really appreciate your input.
Cheers, Adrian

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