How do you solve #9^(x-1)= 5^(2x)#?

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Answer 1 · 2018-03-02T08:58:45

#x=-2.150#

Start by taking the logarithm of both sides, in order to bring the exponents down.

#log_10(9^(x-1))=log_10(5^(2x))#

One of the log rules is this:

#log_a(x^n)=nlog_a(x)#

So:

#(x-1)log_10(9)=2xlog_10(5)#

Evaluate the logarithm and simplify:

#(x-1)(0.9542)=2x(0.6990)#

#0.9542x-0.9542=1.398x#

#-0.9542=0.4438x#

#x=-2.150#