View Full Version : Advanced Functions: Solving Logs
Toaster
Aug 8, 2009, 06:44 PM
Hey guys,
I took a year off and now I'm back in the books.
Now it might be because I've been doing nothing but math today and my brain is fried, but I can't figure out these two problems. Any help is greatly appreciated!
Solve for the variable:
a) logy81=4/3
so I've rewritten it as:
y^4/3= 81
y=?
and that's where I'm at. I have no clue what the base of y is or how to find it.
b)log8 Y=2/3
8^2/3=y
y=?
How am I suppose to write the value for y leave it as 8^2/3?
I hope that wasn't to confusing but thanks for taking a look.
Trevor
galactus
Aug 9, 2009, 03:19 AM
log_{y}(81)=\frac{4}{3}
What this is asking is what number raised to the 4/3 power is equal to 81?
y^{\frac{4}{3}}=81
Try the change of base formula: \frac{log(81)}{log(y)}=\frac{4}{3}
\frac{4log(3)}{3log(3)}=\frac{4}{3}
In other words, what is 3^3 equal?
Unknown008
Aug 9, 2009, 05:45 AM
I don't know if you'll understand my method, but here it is.
log_y (81) = \frac43
y^{\frac43} = 81
But what is 81? It's also 3^4
So,
y^{\frac43} = 3^4
Then 'cube' both sides, then take the forth root of both sides.
y^{\frac43 \times 3} = 3^{4\times 3 }
y^4 = 3^{12}
y^{4 \times \frac14} = 3^{12\times \frac14}
y= 3^{3}
Or simply take the power of 3/4 on both sides.
y^{\frac43 \times \frac34} = 3^{4\times\frac34}
y= 3^{3}
And for the next one, 8 = 2^3
Toaster
Aug 9, 2009, 06:05 AM
Wonderful thanks guys! The problem was I had just forgotten which methods to use, so this help a lot.
@ unknown088, I get what you've done, but what is this formula called?
Thanks again guys!
Unknown008
Aug 9, 2009, 06:09 AM
Huh? Which formula?
These are called logarithmic equations if that was the word you're looking for...
You make use of the power law...
Toaster
Aug 9, 2009, 08:27 AM
Ah k I literally just learned the power law so that makes sense now.
Thanks again!
Toaster
Aug 9, 2009, 05:35 PM
[math]x{\small_1}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/máth]
[math]\normalsize\log_{2}(2m+4)-\log_2(m-1)=3[/máth]
(lets see if this actually works)
galactus
Aug 9, 2009, 05:46 PM
x{\small_1}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\normalsize\log_{2}(2m+4)-\log_2(m-1)=3
(lets see if this actually works)
What is that little thing above the a in [math]? That is why it won't display correctly.
I got rid of it and now it works.
Toaster
Aug 9, 2009, 05:48 PM
Ah thanks galactus! I'll use it next time, by the way I posted a new question if you wouldn't mind taking a look.
Unknown008
Aug 10, 2009, 08:36 AM
LOL! That was the symbol that RickJ posted in his post so that you could see what was the code tags! He specified in his post that you should replace the a symbol by a for the code to work.
EDIT: that was Capuchin, not RickJ, sorry :o
Toaster
Aug 10, 2009, 09:02 AM
Yea what can I say I was extremely tired/didn't read anything just copied and pasted the tags.