How do you solve #(¼)log_6 (a – 3) – log_6 3 = 0#?

1 Answer
Gió
Jun 5, 2016

I found: #a=84#

Explanation:

We can first get rid of #1/4#:
#log_6(a-3)^(1/4)-log_6(3)=0#
Then we change the subtraction operating on the arguments:
#log_6[(a-3)^(1/4)/3]=0#
We use the definition of log:
#(a-3)^(1/4)/3=6^0#
#(a-3)^(1/4)/3=1#
rearrange:
#(a-3)^(1/4)=3#
take the power of #4# on both sides:
#(a-3)^(1/4*4)=3^4#
#a-3=81#
#a=84#

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