Can someone please show me step by step how to solve down on this log problem...

8^-2x+4 = 64

log9x = 1/2

log3 (x-1) + log3 (x+1) = 1

Thank you very much.

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- Apr 8th 2012, 06:21 PM #1

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- Apr 8th 2012, 07:10 PM #2

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## Re: A log problem

I assume the first problem is really 8^(-2x+4)=64

take log_{8}of both sides

log_{8}(8^{-2x+4})=log_{8}(64)

which leads to

(-2x+4)log_{8}(8)=2log_{8}(8)

(-2x+4) = 2

x = 1

second problem

I would first rewrite log_{9}x=1/2 as (ln(x))/(ln(9))=1/2

then

ln(x) = (1/2) ln(9)

ln(x) = ln(9^{1/2})

x = 9^{1/2}=3

third

first rewrite as

(ln(x-1) + ln(x+1))/(ln3)=1

ln(x-1)+ln(x+1) = ln3

ln(x^{2}-1)=ln3

x^{2}-1=3

x^{2}=4

x=2

x can't be -2 because it would cause you to take the log of a negative number

ln=natural log

- Apr 8th 2012, 09:18 PM #3
## Re: A log problem

It's easier to just go

$\displaystyle \displaystyle \begin{align*} \log_9{x} &= \frac{1}{2} \\ x &= 9^{\frac{1}{2}} \\ x &= 3 \end{align*}$

and

$\displaystyle \displaystyle \begin{align*} \log_3{(x - 1)} + \log_3{(x + 1)} &= 1 \\ \log_3{(x - 1)(x + 1)} &= 1 \\ (x - 1)(x + 1) &= 3^1 \\ x^2 - 1 &= 3 \\ x^2 &= 4 \\ x &= \pm 2 \\ x &= 2 \textrm{ due to the fact you can only take the logarithm of a positive number} \end{align*}$