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

1 Answer
Mar 2, 2018

#x=-2.150#

Explanation:

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#