How do you solve #2^(10x+3)=32#?

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Answer 1 · 2016-09-30T10:55:11

#x=1/5#

As #2^((10x+3))=32#

#10x+3=log_2 32#

or #10x+3=log_2 2^5#

or #10x+3=5#

or #10x=5-3=2#

and #x=2/10=1/5#

Answer 2 · 2016-09-30T15:12:39

#x = 1/5#

It is a real advantage to know the powers up to 1000.

Note: #2^5 = 32#

With exponential equations, try to make: Either

The bases equal OR the indices equal

#2^(10x+3) = 32" "# can be written as# " "2^(10x+3) = 2^5#

THerefore: #10x+3 = 5" "larr# bases are equal, indices are equal

#10 x = 5-3#
#10x = 2#

#x = 10/2#
#x = 1/5#