Question

How do you solve `Log_2 (x-4) + log_2 (x+4) = 3`?

Answer 1

Use sum of the logarithms rule and condense the logarithms on the left. Covert logarithms to exponent form to solve. The steps are given below.

Explanation:

`log_2(x-4)+log_2(x+4)=3`

Let us use the rule `log(A) + log(B) = log(AB)`

`log_2(x-4)(x+4)=3`

`log_2(x^2-16)=3` Note : `(a-b)(a+b) = a^2 - b^2`

Converting the logarithms to exponent form

If `log_b(a) =k` then `a=b^k`

`log_2(x^2-16) = 3`
`x^2-16 = 2^3`
`x^2-16 = 8`
`x^2-16+16 = 8+16`
`x^2=24`
take square root on both the sides.
`x = +-sqrt(24)`
`x=+-sqrt(4*6)`
`x=+-sqrt(4)*sqrt(6)`
`x=+-2sqrt(6)`

`x=2sqrt(6) or x = -2sqrt(6)`

Check the validity of solution by substituting the solution in the original equation.

We can see for `log_2(x-4)` if we substitute `x=-2sqrt(6)` the logarithm is not defined.

Therefore the final answer is `x=2sqrt(6)`