algebra 2 problem

Question

8x-21. 8^-x =4

 

Could somebody give me the answer to this please

Answer 1

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

 

8^x-21(1/8^x)=4

 

Multiply both sides by 8^x:  8^2x-4(8^x)-21 = 0

 

Let u = 8^x to obtain u² - 4u - 21 = 0

 

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

 

u = 7 or u = -3

 

So, 8^x = 7 or 8^x = -3

 

8^x>0 for all x, so the equation 8^x = -3 has no solution.

 

Thus, 8^x = 7

 

Taking logs of both  sides, we get log(8^x) = log7

 

                                                x(log8) = log7

 

                                                       x = (log7)/(log8)

 

                                                          ≈ 0.94

 

                                                          

 

                                            

Answer 2

Hi Debbie,

 

If this is exponential or logarithmic function (if both Xs are exponents), here are both exact and decimal answer:

8^x - 21.8^-x = 4

I took log both sides:

log (8^x - 21.8^-x) = log 4

log 8^x - log 21.8^-x = log 4

Power (exponent) comes to the front

x log 8 - (- x) log 21.8 = log 4 (property of logarithm)

x log 8 + x log 21.8 = log 4

 

x (log 8 + log 21.8) = log 4        (Let's evaluate what's inside the parentheses)

log _z (ab) = log _z a + log z b       (property of logarithm)

log (8 * 21.8) = log 8 + log 21.8 (property of logarithm)

log (174.4) = log 8 + log 21.8

 

x (log 8 + log 21.8)   =   log 4

x (log 174.4) = log 4

x = (log 4) / (log 174.4)                             exact answer

x = 0.60206 / 2.2415 = 0.2685969             decimal answer