How do you solve #25^ { x } - 8\cdot 5^ { x } = - 16#?

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Answer 1 · 2018-06-22T07:00:11

#x=log_5(2)#

#(5^2)^x-8 *5^x=-16#
#t^2-8t=-16#
#t=4#
#5^x=4#

Answer 2 · 2018-06-22T07:01:01

#color(maroon)(x = log 4 / log 5 = 0.8614#

#25^x - 8 * 5^x + 16 = 0#

Let #5^x = a#

#=> a^2 - 8a + 4^2 = 0#

#(a-4)^2 = 0#

#:. a = 4#

#a = 5^x = 4#

#x log 5 = log 4#

#x = log(4) / log(5) = 0.8614#