Question

How do you solve these logarithmic equations?

`4+log_(9)(3x-7)=6`
`log_(2)(2x)+log_(2)x=5`
`3log_(5)x-log_(5)(5x)=3-log_(5)25`

Answer 1

See below.

Explanation:

Before the logarithm application, the equations read

`9^4(3x-7)=9^6`
`(2x)x=2^5`
`x^3/(5x)=5^3/25`

after that we have

`3x-7=9^2->x=(9^2+7)/3`
`x^2=2^4->x=pm2^2`
`x^2=5^2->x=pm 5`

Answer 2

`4+log_9(3x-7)=6`

`log_9(3x-7)=2`

`3x-7=81`

`3x=89`

`x=89/3`

`log_2(2x)+log_2(x)=5`

`log_2(2x^2)=5`

`2x^2=2^5=32`

`x^2=16`

`x=+-4`

`3log_(5)x-log_(5)(5x)=3-log_(5)25`

`log_5x^3-log_5(5x)=3-2=1`

`log_5(x^3/(5x))=log_5(x^2/5)=1`

`x^2/5=5`

`x^2=25`

`x=+-5`