If #5^(2x) + 3(5)^x = 28#, then what is the value of #x#?
1 Answer
Dec 7, 2016
Let
#t^2 + 3t = 28#
#t^2 + 3t - 28 =0#
#(t + 7)(t - 4) = 0#
#t= -7 and 4#
Now, since
#5^x = -7 and 5^x = 4#
#ln(5^x) = ln(-7) and ln(5^x) = ln4#
#x = O/ and xln5 = ln4#
#x = ln4/ln5#
Hopefully this helps!
