1. A log problem

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.

2. Re: A log problem

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

take log8 of both sides

log8(8-2x+4)=log8(64)

(-2x+4)log8(8)=2log8(8)

(-2x+4) = 2

x = 1

second problem

I would first rewrite log9x=1/2 as (ln(x))/(ln(9))=1/2

then

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

ln(x) = ln(91/2)

x = 91/2=3

third

first rewrite as

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

ln(x-1)+ln(x+1) = ln3
ln(x2-1)=ln3
x2-1=3
x2=4
x=2

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

ln=natural log

3. Re: A log problem

Originally Posted by bmoon123
second problem

I would first rewrite log9x=1/2 as (ln(x))/(ln(9))=1/2

then

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

ln(x) = ln(91/2)

x = 91/2=3

third

first rewrite as

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

ln(x-1)+ln(x+1) = ln3
ln(x2-1)=ln3
x2-1=3
x2=4
x=2

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

ln=natural log
It's easier to just go

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

and

\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*}