1. ## Logarithms, Algebra 2

You last visited: 09-17-2007 at 06:58 PM

Sorry...the last time I visited here was when I was in like algebra 1, lol.
I need help with some math problems about logs (I have a math test tomorrow ), I'd really appreciate the help. I'm sorry, I don't know how to make it look like the problems on my paper, but I hope this is understandable:

log[base 5](1/8) / log[base 5](2)

log
base 5
one-eighth
divided by
log
base 5
two

I think this would go along with log[base 5](1/8) minus log[base5](2) but I can't figure it out

Here is another:

7^log[base7](16) minus log[base 7](2) = x

I was absent both the above one was done and the one below:

log[base 7](log[base 3](log[base 2](x))) = 0

They are both making me

There is one last one (Sorry for the amount...I know a few of my friends can tell me how to do some in my free time before the test, but if I can get as much done here as possible, that would be awesome.)

log (x^2) = (log x)^2

Those logs above would be base 10

If anyone could help I'll give you free cookies, thanks!

2. Hello, Harryhit4!

$\frac{\log_5\left(\tfrac{1}{8}\right)} {\log_5(2)}$
There is a base-change formula: . $\log_a(x) \;=\;\frac{\log_b(x)}{\log_b(a)}$

Reading it backwards, we have: . $\frac{\log_5\!\left(\frac{1}{8}\right)}{\log_5(2)} \;=\;\log_2\!\left(\frac{1}{8}\right)$

Then: . $\log_2\!\left(\frac{1}{8}\right) \;=\;\log_2\left(2^{-3}\right) \;=\;-3\cdot\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\boxed{-3}$

$7^{\log_7(16)} - \log_7(2) \:=\:x$
This is a strange one . . . Is there a typo?

We have: . $16 - \log_7(2) \:=\:x\quad\Rightarrow\quad \boxed{x \:=\:16-\log_7(2)}$

$\log_7\bigg(\log_3\left[\log_2(x)\right]\bigg) \:=\: 0$
I hope you can follow these steps . . .

$\log_3\bigg(\log_2(x)\bigg) \:=\:7^0 \quad\Rightarrow\quad \log_3\bigg(\log_2(x)\bigg) \:=\:1$

$\log_2(x) \:=\:3^1 \quad\Rightarrow\quad \log_2(x) \:=\:3$

$x \:=\:2^3 \quad\Rightarrow\quad\boxed{ x \:=\:8}$

$\log\left(x^2\right) \:= \:\left(\log x\right)^2$

We have: . $2\!\cdot\!\log x \:=\:\left(\log x\right)^2 \quad\Rightarrow\quad \left(\log x\right)^2 - 2\!\cdot\!\log x \:=\:0$

Factor: . $\log x\,\left(\log x - 2\right) \:=\:0$

. . We have: . $\log x \:=\:0 \quad\Rightarrow\quad x \:=\:10^0 \:=\:\boxed{1}$

. . . And: . $\log x - 2 \:=\:0 \quad\Rightarrow\quad \log x \:=\:2\quad\Rightarrow\quad x \:=\:10^2 \:=\:\boxed{100}$

3. Originally Posted by Soroban
Hello, Harryhit4!

There is a base-change formula: . $\log_a(x) \;=\;\frac{\log_b(x)}{\log_b(a)}$

Reading it backwards, we have: . $\frac{\log_5\!\left(\frac{1}{8}\right)}{\log_5(2)} \;=\;\log_2\!\left(\frac{1}{8}\right)$

Then: . $\log_2\!\left(\frac{1}{8}\right) \;=\;\log_2\left(2^{-3}\right) \;=\;-3\cdot\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\boxed{-3}$

This is a strange one . . . Is there a typo?

We have: . $16 - \log_7(2) \:=\:x\quad\Rightarrow\quad \boxed{x \:=\:16-\log_7(2)}$

I hope you can follow these steps . . .

$\log_3\bigg(\log_2(x)\bigg) \:=\:7^0 \quad\Rightarrow\quad \log_3\bigg(\log_2(x)\bigg) \:=\:1$

$\log_2(x) \:=\:3^1 \quad\Rightarrow\quad \log_2(x) \:=\:3$

$x \:=\:2^3 \quad\Rightarrow\quad\boxed{ x \:=\:8}$

We have: . $2\!\cdot\!\log x \:=\:\left(\log x\right)^2 \quad\Rightarrow\quad \left(\log x\right)^2 - 2\!\cdot\!\log x \:=\:0$

Factor: . $\log x\,\left(\log x - 2\right) \:=\:0$

. . We have: . $\log x \:=\:0 \quad\Rightarrow\quad x \:=\:10^0 \:=\:\boxed{1}$

. . . And: . $\log x - 2 \:=\:0 \quad\Rightarrow\quad \log x \:=\:2\quad\Rightarrow\quad x \:=\:10^2 \:=\:\boxed{100}$

Thank you very, very much for your help! I understand them all...

The second one - the minus log 7 was still supposed to be in the exponent of 7. Sorry for the confusion. You probably won't be able to reply by the time I leave for school (15 mins) but I will ask my friends or my teacher. Thanks!