How do you solve #4^(2x-5)=32#?

Note that #4 = 2^2# and #32 = 2^5#.

So we have:

#2^(4x-10) = (2^2)^(2x-5) = 4^(2x-5) = 32 = 2^5#

The function #2^x# is one to one as a real valued function from #RR# to #(0, oo)#.

Hence the only real solution is given by:

#4x-10 = 5#

Hence:

#x = 15/4#

If we are interested in other complex solutions, note that:

#2^((2kpii)/ln 2) = e^(2kpii) = (e^(2pii))^k = 1^k = 1" "# for any integer #k#

Hence the given equation has solutions given by:

#4x-10 = 5+(2kpii)/ln 2#

Hence:

#x = 15/4+(kpi)/(2 ln 2) i#

In order to solve, you can manipulate the bases of the exponents to find a value for x.

In this question, you can manipulate the base of 4 to be #2^2# and 32 to be #2^5#. It should look like this:
#2^(2(2x-5))=2^5#

Now that the bases equivalent, you can take a logarithm with a base of 2 to both sides of the equation to cancel out the exponents.
#log_2(2^(2(2x-5)))=log_2(2^5)#

#log_2(2)=1# so you can cancel out the log functions on both sides. This gives you:
#2(2x-5)=5#

Now, just solve like a normal algebraic equation.
#4x-10=5#
#4x=15#
#x=15/4#