# Thread: Need help with log question?

1. ## Need help with log question?

I was doing practice exams and I got stuck on this question?

Solve for x in the equation: 2a^x = b^(x+1)

I tried it over and over and still can't get x by itself, how do you do this question? It's on the no calculator section.

The answer's x = (log2 - log b) / (log b - log a)

If it helps.

2. Are you sure the answer is correct? Anyway, I will show you my method and you compare it.
$\displaystyle 2a^x=b^(x+1)$
then, $\displaystyle (x)\lg{2a} = (x+1)(\lg{b})$
After which, $\displaystyle x(\lg{2}+\lg{a}) = (x)\lg{b} + \lg{b}$
Finally, $\displaystyle x(\lg{2} + \lg{a} - \lg{b}) = \lg{b}$
You arrive at the answer, $\displaystyle x = \frac{\lg{b}}{\lg{2} + \lg{a} - \lg{b}}$
This may be wrong.

3. Hello, imperfections!

Solve for $\displaystyle x\!:\;\;2a^x \:=\: b^{x+1}$

Answer: .$\displaystyle x \:= \:\frac{\log2 - \log b}{\log b - \log a}$

Take logs of both sides: .$\displaystyle \log\left(2a^x\right) \:=\:\log\left(b^{x+1}\right)$

. . . . . . .$\displaystyle \log2 + \log(a^x) \:=\:\log(b^{x+1})$

. . . . . . .$\displaystyle \log2 + x\!\cdot\log a \:=\:(x+1)\log b$

. . . . . . .$\displaystyle \log2 + x\!\cdot\log a \:=\:x\!\cdot\!\log b + \log b$

. . . . . .$\displaystyle x\!\cdot\!\log a - x\!\cdot\!\log b \:=\:\log b - \log 2$

Factor: .$\displaystyle x\!\cdot\!(\log a - \log b) \:=\:\log b - \log 2$

Therefore: .$\displaystyle \boxed{x \;=\;\frac{\log b - \log 2}{\log a - \log b}}$

To get their answer, multiply by $\displaystyle \tfrac{\text{-}1}{\text{-}1}$

. . $\displaystyle x \;=\;\frac{\text{-}1}{\text{-}1}\cdot\frac{\log b - \log 2}{\log a - \log b} \;=\;\frac{\log 2 - \log b}{\log b - \log a}$