What is Solution

Question

log₃(3^x-8)=(2-x)

Answer 1

Log₃(3^x - 8) = (2-x) => 3^(2-x) = (3^x -8) Then (3² /3^x ) = (3^x - 8) ; Multiplying by 3^x  in both sides, we obtain 3² = (3^(2x) - 8*3^x ) => This needs to be solved by trial and error

Answer 2

First of all, take both side to the power of 3.

 

3^{log₃(3^x-8)}  = 3^(2-x)

 

Next, simplify:

 

(3^x  - 8) = 3^(2-x)

 

Use property of exponents to expand the left side (x^m/xⁿ⁾= x^(m-n)

 

3^x - 8 = 3²/3^x

 

Multiply both sides by 3^x.

 

Enjoy!

 

Jeff Negus

 

3^2x - 8*3^x = 9

 

Subtract 9 from each side:

 

3^2x - 8 * 3^x  - 9 = 0

 

Let y = 3^x

 

y² - 8y - 9 = 0

(y-9)(y+1) = 0

3^x = 9 OR 3^x = -1

x = 2 since 3^x = -1 is impossible.

 

Check:

 

log₃(3² - 8) = (2 - 2)

 

log₃(9 - 8) = 0

 

log₃(1) = 0

 

0 = 0 √