SOLUTION: Solve 8^(x+8) = 32^x
The lesson this is found is called Exponential and Logarithmic Equations
I am looking for more of the steps on how to complete rather then the answer but bo
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Solve 8^(x+8) = 32^x
The lesson this is found is called Exponential and Logarithmic Equations
I am looking for more of the steps on how to complete rather then the answer but bo
Log On
Question 596172: Solve 8^(x+8) = 32^x
The lesson this is found is called Exponential and Logarithmic Equations
I am looking for more of the steps on how to complete rather then the answer but both would be great. Thanks for your help
The steps I have tried are
8 ^(x+8) = 32^x ( Original Problem)
log 8^(x+8)= log 32^x
(x+8) log 8 = x log 32
(log8)x +8log8 = (log 32)x
8log8 = x log 32- x log 8
8log8 = x log 24 Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The steps I have tried are
8 ^(x+8) = 32^x ( Original Problem)
log 8^(x+8)= log 32^x
(x+8) log 8 = x log 32
(log8)x + 8log8 = (log 32)x
8log8 = x log 32- x log 8
8log8 = x log 24 ********* That's not right
---------------------------
8log8 = x log 32- x log 8
8log8 = x*(log 32 - log 8)
8lob(8) = x*log(4) Subtracting logs is division
x = 8log(8)/log(4)
x = 12
-----------------
A better way:
8 ^(x+8) = 32^x
8 = 2^3 and 32 = 2^5
3x+24 = 5x
2x + 24
x = 12
