Question 596172
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
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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
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A better way:
8 ^(x+8) = 32^x
8 = 2^3 and 32 = 2^5
{{{2^(3x+24) = 2^(5x)}}}
3x+24 = 5x
2x + 24
x = 12