algebra 2 question

Question

8^x -21. 8^-x =4

 

 

what is exact solution and decimal approximation

Answer 1

Rewrite the equation as:

8^x - 21/8^x = 4

Multiply every term by 8^x.

8^2x - 21 = 4(8^x)

Rewrite:

(8^x)² - 4(8^x) - 21 = 0

This is quadratic in 8^x. Let u = 8^x.

u² - 4u - 21 = 0

(u - 7) (u + 3) = 0

u = 7 OR u = -3

So:

8^x = 7 OR 8^x = -3.

There is no value of x such that 8^x = -3, so focus on 8^x = 7 (taking the natural log of both sides):

ln 8^x = ln 7

x ln 8 = ln 7

x = ln7 / ln 8 ≈ 0.93578

Note that ln 7 / ln 8 is the exact solution. This could also be written as log₈ 7.