Question

Solve for x (log question)?

`log_7 x-1/2log_7 4=1/2log_7 (2x-3)`

Answer 1

`x=2,6`

Explanation:

Problem: `log_7(x)-1/2log_7(4)=1/2log_7(2x-3)`

We can use the logarithm power rule

`alog_c(b)=log_c(b^a)`

on the right side of the equation, which will give us

`log_7(x)-log_7(4)^(1/2)=log_7(2x-3)^(1/2)`

`log_7(x)-log_7sqrt(4)=log_7sqrt(2x-3)`

Because logarithms only work for positive numbers, then `sqrt(4)=2`.

`log_7(x)-log_7(2)=log_7sqrt(2x-3)`

Adding `log_7(2)` to both sides, we get

`log_7(x)=log_7sqrt(2x-3)+log_7(2)`

Use the logarithm property `log_x(a)+log_x(b)=log_x(ab)`,

`log_7(x)=log_7(2sqrt(2x-3))`

`x = 2sqrt(2x-3)`

Now, we can solve for `x`.

Square both sides to get

`x^2=4(2x-3)`

`x^2=8x-12`

`x^2-8x+12=0`

`(x-6)(x-2)=0`

`:.x=2,6`

Answer 2

`x= 6 or 2`

Explanation:

`log_7 x - 1/2log_7 4 = 1/2log_7 (2x-3)`

Remember: `mlog_n a = log_n a^m`

`:.log_7 x - log_7 4^(1/2) = log_7 (2x-3)^(1/2)`

Also remember: `log_n a - log_n b = log_n (a/b)`

`:. log_7 (x/(sqrt4)) = log_7 (sqrt(2x-3))`

Finally remember: `log_n a = log_n b -> a=b`

`:. x/2 = sqrt(2x-3)`

Square both sides.

`x^2/4 = 2x-3`

`x^2-8x+12=0`

`(x-6)(x-2) =0`

Hence, `x=6 or 2`