Question

How do you solve `log_4 x^2 - log_4 (x+1) = 5`?

Answer 1

`x ~~ - 0.999 " or " x ~~ 1024.999`

Explanation:

1) Determine the domain

First of all, let's determin the domain of the two logarithmic expressions.

• As `x^2 >= 0` for all `x in RR`, `log_4(x^2)` is defined for all `x != 0`.
• `log_4(x+1)` is defined for `x + 1 > 0 <=> x > -1`

Thus, our domain is `x > -1` and `x != 0`.

2) Solve the equation

Now let's solve the equation. To start, use the logarithmic law

`log_a (m) - log_a(n) = log_a(m/n)`

Thus, you can transform your equation into:

`log_4(x^2 / (x+1)) = 5`

Now, the inverse function of `log_4(x)` is `4^x` which means that both `log_4(4^x) = x` and `4^(log_4(x)) = x` hold.

Thus, to "eliminate" the logarithmic term, you need to apply `4^x` to both sides of the equation:

`x^2/(x+1) = 4^5`

`x^2/(x+1) = 1024`

... multiply both sides with `(x+1)`...

`x^2 = 1024(x+1)`

... bring all the terms to the left side...

`x^2 - 1024x - 1024 = 0`

Solve the equation e.g. with the quadratic formula:

`x = (1024 +- sqrt(1024^2 + 4 * 1024))/2 ~~ (1024 +- 1025.998)/2`

`x ~~ - 0.999 " or " x ~~ 1024.999`

3) Check the domain

Now, we need to check if both `x` fit into our domain.

Indeed, `x > -1` and `x != 0` holds for both of them, so both are solutions.