Question 557415: log8x+log64(x+2)=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! log8x+log64(x+2)=1
log8x+log64+log(x+2)=1
log8x+log(x+2)=1-log64=-0.80617997
log[(8x)(x+2)]=-0.80617997
convert to exponential form
10^-0.80617997=8x*(x+2)
0.15625=8x^2+16x
8x^2+16x-.15625=0
solve by quadratic formula:
x=-2.00972(reject, x>0)
or
x=0.0097184
Check:
log 8x=-1.10931524
log64(x+2)=log(64*2.0097184)=2.10931518
log8x+log64(x+2)=-1.10931524+2.10931518=0.99999994≈1
anybody know an easier way?
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