How do you solve # 1/2log_a(x + 2) + 1/2log_a(x - 1) = 2/3log_a27#?

1 Answer
Gió
Dec 4, 2015

I found: #x=(-1+3sqrt(37))/2#

Explanation:

We can try some manipulations:
collect #1/2#:
#1/2[log_a(x+2)+log_a(x-1)]=2/3log_a27#
#log_a(x+2)+log_a(x-1)=2*2/3log_a27#

use the property of logs that says:
#logx+logy=log(xy)#
#log_a[(x+2)(x-1)]=4/3log_a27#

use the other property of logs that says:
#alogx=logx^a#
#log_a[(x+2)(x-1)]=log_a27^(4/3)#

if the two logs must be equal then the argumens must be as well:
#(x+2)(x-1)=27^(4/3)#
#(x+2)(x-1)=root3(27^4)#
#(x+2)(x-1)=81#
#x^2-x+2x-2-81=0#
#x^2+x-83=0#

You can now use the Quadratic Formula to get:
#x_(1,2)=(-1+-sqrt(1+332))/2=#
two solutions:
#x_1=(-1-3sqrt(37))/2# NO it will give you a negative log.
#x_2=(-1+3sqrt(37))/2#

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