A log problem I'm not sure where to start with

[maccy]

Given:

8^-2x+4 = 64

log9x = 1/2

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

Could someone show me step by step how to solve this problem? Would be appreciated.

Thanks for your amazing services

 

[Deleted member 4993]

Given:

8^-2x+4 = 64

log9x = 1/2

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

Could someone show me step by step how to solve this problem? Would be appreciated.

Thanks for your amazing services

You have three problems here.

For the second and third problem - are the bases of log function assumed to be 10?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[soroban]

Hello, maccy!

$\displaystyle 1.\;8^{-2x+4} \:=\: 64$

Get the same base on both sides: .$\displaystyle 8^{-2x+4} \:=\:8^2$

Equate exponents: .\(\displaystyle -2x + 4 \:=\:2 \quad\Rightarrow\quad -2x \:=\:-2 \quad\Rightaffow\quad x \:=\:1\)

$\displaystyle 2.\;\log_9x \:=\: \frac{1}{2}$

Write in exponential form: .$\displaystyle x \:=\:9^{\frac{1}{2}} \quad\Rightarrow\quad x \,=\,3$

$\displaystyle 3.\;\log_3(x-1)+\log_3(x+1) \:=\: 1$

Combine the logs: .$\displaystyle \log_3(x-1)(x+1) \:=\:1\quad\Rightarrow\quad \log_3(x^2-1) \:=\:1$

Write in exponential form: .$\displaystyle x^2-1 \:=\:3^1 \quad\Rightarrow\quad x^2 \:=\:4 \quad\Rightarrow\quad x \:=\:2,\;\rlap{//}\text{-}2$