Solving Exponential and Logarithmic Equations

[megslove1212]

I need help explaining how to do this type of problem.

log3(x-9)+ log3(x-3)=2

and

3^(x+4)= 6^(2x-5)

Help please! THANKS!

 

[galactus]

I need help explaining how to do this type of problem.

log3(x-9)+ log3(x-3)=2

Log law:

\(\displaystyle log_{3}(x-9)+log_{3}(x-3)=log_{3}((x-9)(x-3))\)

\(\displaystyle log_{3}((x-9)(x-3))=2\)

base rule:

\(\displaystyle (x-9)(x-3)=3^{2}\)

Now, solve the quadratic. Make sure to check both solutions. Only one may be valid.

3^(x+4)= 6^(2x-5)

Help please! THANKS!

\(\displaystyle 3^{x+4}=6^{2x-5}\)

\(\displaystyle 3^{x}\cdot 81=\frac{36^{x}}{7776}\)

\(\displaystyle 3^{x}\cdot 629856=36^{x}\)

\(\displaystyle 629856=12^{x}\)

12 to what power equals 629856?.

 

[lookagain]

Now, solve the quadratic. Make sure to check both solutions. Only one may be valid.

Or no solutions may be valid.