Asked by Jake
Suppose 81^x=64. What is 27^(x+1)?
Answers
Answered by
Damon
81 = 3^4 so 3^4x = 64
27 = 3^3 so we want 3^(3x+3)
3^4x = 2^6
4x log 3 = 6 log 2
x = 6 log2 /4 log 3 = 1.5 log2/log3
= whatever, calculator
so
y = 3^(3x+3)
log y = (3x+3) log 3
find log y
take 10^log y = y
27 = 3^3 so we want 3^(3x+3)
3^4x = 2^6
4x log 3 = 6 log 2
x = 6 log2 /4 log 3 = 1.5 log2/log3
= whatever, calculator
so
y = 3^(3x+3)
log y = (3x+3) log 3
find log y
take 10^log y = y
Answered by
Reiny
81^x = 64
log 81^x = log 64
x log81 = log64
x = log64/log81
27^(log64/log81 + 1)
= 610.940...
log 81^x = log 64
x log81 = log64
x = log64/log81
27^(log64/log81 + 1)
= 610.940...
Answered by
Jake
Our teacher hasn't taught us log, so how would I sove this problem?
Answered by
Damon
I do not know. Make sure you have no typos. I think it was meant to be easier.
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