lnx−4ln3=ln(5x)
Recall that alnb=ln(ab), so
lnx−ln34=ln(5x)
Recall that lna−lnb=ln(ab), so
ln(x34)=ln(5x)
Convert the logarithmic equation to an exponential equation.
eln(x34)=eln(5x)
Remember that elnx=x, so
x34=5x
x2=5×34
x=±√5×34=±√5×√34=±√5×32
x=9√5 and x=−9√5 are possible solutions.
Check:
lnx−4ln3=ln(5x)
If x=9√5,
ln(9√5)−4ln3=ln(59√5)
ln9+ln√5−ln34=ln5−ln(9√5)
ln9+ln√5−ln(32)2=ln(√5)2−ln9−ln√5
ln9+ln√5−2ln9=2ln√5−ln9−ln√5
ln√5−ln9=ln√5−ln9
∴ x=9√5 is a solution.
If x=−9√5,
ln(−9√5)−4ln3=ln(5−9√5)=ln5−ln(−9√5)
But ln(−9√5) is not defined.
∴ x=−9√5 is not a solution.
