How do you solve #7^(x - 2) = 5^x#?

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Answer 1 · 2016-02-24T14:56:12

Convert to logarithmic form.

#log7^(x - 2) = log5^x#

#(x - 2)log7 = xlog5#

#xlog7 - 2log7.=x log 5#

#xlog7 - xlog5 = 2log7#

#x(log7 - log5) = 2log7#

#x = (2log7)/(log7 - log5)#

You can simplify the answer further by using the rules #alogn = logn^a# and #log_an - log_am = log_a(n / m)#

#x = log7^2/log(7/5)#

#x = log49/log(7/5)#

Hopefully this helps!