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3 answers

log5^x=log32
xlog5=log32
x=log32/log5
x=2.15

2007-11-28 09:35:37 · answer #1 · answered by Anonymous · 0 0

xlog 5 = log 32
x = log 32/log 5
=2.15

2007-11-28 09:36:31 · answer #2 · answered by norman 7 · 0 0

You have to use logarithms. log5^x=log32 (take the log of both sides), and then that equals x*log5=log32, which is
x=log32/log5

2007-11-28 09:36:41 · answer #3 · answered by Sophie 1 · 0 0

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