# Thread: Question with Logarithms

1. ## Question with Logarithms

How do I turn $\displaystyle 3^x$ = $\displaystyle 4^x^-^2$ into $\displaystyle \frac{\log\frac1{16}}{log\frac3{4}}$ and could you please also explain the steps thoroughly. Thanks!

2. ## Re: Question with Logarithms

1st, you don't. The first is an EQUATION and the second is an EXPRESSION. Not the same at all.

2nd, this is a logarithm problem, as you can see by the final form. If everything is suitable, then $\displaystyle a^{b} = c^{d}$ implies $\displaystyle b\cdot log(a) = d\cdot log(c)$. That's almost all you need.

3. ## Re: Question with Logarithms

OH sorry i completely forgot to make the second equal to x.

4. ## Re: Question with Logarithms

Originally Posted by Jayissocool
How do I turn $\displaystyle 3^x$ = $\displaystyle 4^x^-^2$ into $\displaystyle \frac{\log\frac1{16}}{log\frac3{4}}$ and could you please also explain the steps thoroughly. Thanks!
I'll show you how to start and leave the rest for you:

$\displaystyle 3^x = 4^{x-2}$

$\displaystyle 3^x = 4^x \cdot 4^{-2}$

$\displaystyle \frac{3^x}{4^x}=\frac1{16}$

$\displaystyle \left(\frac34 \right)^x=\frac1{16}$

... and now it's your turn.

5. ## Re: Question with Logarithms

Hello, Jayissocool!

First of all, that is a really stupid way to leave the answer!

$\displaystyle \text{How do I turn }\,3^x \:=\:4^x^{-2}\,\text{ into }\,x \:=\:\frac{\log\frac1{16}}{\log\frac3{4}}\;?$

Take logs: .$\displaystyle \log\left(3^x) \;=\;\log\left(4^{x-2}\right) \quad\Rightarrow\quad x\log3 \;=\;(x-2)\log4$

. . $\displaystyle x\log 3 \;=\;x\log4 - 2\log 4 \quad\Rightarrow\quad x\log 3 - x\log4 \;=\;-2\log 4$

. . $\displaystyle x(\log3-\log4) \;=\;\log(4^{-2}) \quad\Rightarrow\quad x \;=\;\frac{\log(4^{-2})}{\log3-\log 4}$

Therefore: .$\displaystyle x \;=\;\dfrac{\log\frac{1}{16}}{\log\frac{3}{4}} \;\;\hdots\text{ which equals: }\,x \,=\,\frac{\log16}{\log\frac{4}{3}}$