We can try some manipulations:
collect 1/212:
1/2[log_a(x+2)+log_a(x-1)]=2/3log_a2712[loga(x+2)+loga(x−1)]=23loga27
log_a(x+2)+log_a(x-1)=2*2/3log_a27loga(x+2)+loga(x−1)=2⋅23loga27
use the property of logs that says:
logx+logy=log(xy)logx+logy=log(xy)
log_a[(x+2)(x-1)]=4/3log_a27loga[(x+2)(x−1)]=43loga27
use the other property of logs that says:
alogx=logx^aalogx=logxa
log_a[(x+2)(x-1)]=log_a27^(4/3)loga[(x+2)(x−1)]=loga2743
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)=2743
(x+2)(x-1)=root3(27^4)(x+2)(x−1)=3√274
(x+2)(x-1)=81(x+2)(x−1)=81
x^2-x+2x-2-81=0x2−x+2x−2−81=0
x^2+x-83=0x2+x−83=0
You can now use the Quadratic Formula to get:
x_(1,2)=(-1+-sqrt(1+332))/2=x1,2=−1±√1+3322=
two solutions:
x_1=(-1-3sqrt(37))/2x1=−1−3√372 NO it will give you a negative log.
x_2=(-1+3sqrt(37))/2x2=−1+3√372
