How do you solve #2^x = 5^(x - 2)#?

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Answer 1 · 2016-04-11T00:02:21

Since #a = b -> loga = logb#

#log2^x = log5^(x - 2)#

Use the rule #loga^n = nloga#

#xlog2 = (x - 2)log5#

#xlog2 = xlog5 - 2log5#

#xlog2 - xlog5 = -2log5#

#x(log(2/5))= log(1/25)#

#x = (log(1/25))/(log(2/5))#

That's your answer.

Hopefully that helps!