How do you solve #5^ { x + 2} = 5^ { 9} #?

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Answer 1 · 2017-05-06T03:18:56

#x= 7#

When the base number on the LEFT-hand side equals to the base number on the RIGHT-hand side, we can simply equate their exponents/power.

#5^(x+2) = 5^9#

By equating the exponents on the LHS and RHS, we get:

#x+2 = 9#
#color(red)(x= 7)#

Double checking the solution,

#5^(7+2) = 5^9#
#5^9 = 5^9# Hurray!

Alternatively , we can use the logarithm to solve.
By adding #log_5# to both sides, we get:

#log_5 5^(x+2) = log_5 5^9#

The special properties of logarithmic functions allow us to "bring" down the exponent as such:

#(x+2) log_5 5 = 9log_5 5#

When the base b, of the logarithmic functions is the same as the number, #log_b b = 1#.

Therefore, #(x+2)# x #1 = 9# x #1#

#x+2 = 9#
#color(red)(x=7)#

Same as the previous answer!