SOLUTION: log base4(7x-8)=log base4(5x+3)
Algebra.Com
Question 605353: log base4(7x-8)=log base4(5x+3)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
log base4(7x-8)=log base4(5x+3)
log4(7x-8)-log4(5x+3)=0
place under single log
log[(7x-8)/(5x+3)]=0
convert to exponential form: base(4) raised to log of number(0)=number[(7x-8)/(5x+3)]
4^0=[(7x-8)/(5x+3)]=1
7x-8=5x+3
2x=11
x=5.5
RELATED QUESTIONS
Solve: 2log(base4)x+log(base4)3=log(base4)x-log(base4)2
(answered by rapaljer)
Solve for x:
log base4 (x) + log base 4 (x+3) = log base4... (answered by lwsshak3)
log(base4)(x-3)+log(base4)(x+9)=3 (answered by tommyt3rd,josmiceli)
Log base4 (x+5) = log base4... (answered by jim_thompson5910)
log(base4)(x-9)-log(base4)(x+3)=2 (answered by lwsshak3)
Log(base4)(2x+1)=Log(base4)(x-3)+Log(base4)(x+5) (answered by Nate)
(1/3)log(base4)*27-(2log(base4)*6-(1/2)log(base4)*81 (answered by lwsshak3)
Log base4 (x^2 - 3) + Lobg base4... (answered by Earlsdon)
Simplify: 2log(base4)9 -... (answered by stanbon)