How do you solve #log_x 6 = 0.5#?

1 Answer
Laura
Jan 8, 2016

You solve any simple log equation by putting the equation back into exponential form.

Explanation:

Essentially this question is asking you base #x# to exponent #1/2# is 6.

#log_x6=0.5#
To put this into exponential form, write it like #x^0.5=6#.
Which is #sqrtx=6#
The inverse operation of square root is #x^2#, so simply square both sides to get your answer.

#(sqrtx)^2=6^2#
#x=36#

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