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How do you solve #25^x = 5^(x^2 - 15)#?

1 Answer

Mar 28, 2016

#x={-3,5}#

Explanation:

#25^x=5^(x^2-15)#

#color(red)(5)^(2x)=color(red)(5)^(x^2-15)#
#"equation has the same base value"#

#2x=x^2-15#

#x^2-2x-15=0#

#(x-5)(x+3)=0#
#if (x-5)=0" then " x=5#
#if(x+3)=0" then "x=-3#
#x={-3,5}#

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