Question

Solve:

`log_5(24-x)+log_5(-x)=2`

Answer 1

x = -1

Explanation:

Given equation:

`\log_5(24-x)+\log_5(-x)=2`

`\textsf{Apply the log Product law}: \quad \log_ax + \log_ay=\log_axy`

`\implies \log_5[-x(24-x)]=2`

`\implies \log_5(x^2-24x)=2`

`\textsf{Apply the log law}: \quad \log_ab=c \iff a^c=b`

`\implies 5^2=x^2-24x`

Simplify:

`\implies x^2-24x-25=0`

Factor the quadratic equation:

`\implies x^2+x-25x-25=0`

`\implies x(x+1)-25(x+1)=0`

`\implies (x-25)(x+1)=0`

Apply the zero product property:

`\implies x-25=0 \implies x=25`

`\implies x+1=0 \implies x=-1`

As logs of negative numbers cannot be taken, x = -1 is the only valid solution.

Check by substituting x = -1 into the original equation:

`\implies \log_5(24-(-1))+\log_5(-(-1)`

`\implies \log_5(25)+\log_5(1)`

`\implies \log_5(5^2)+0`

`\implies 2\log_5(5)`

`\implies 2(1)`

`\implies 2`

Hence, the solution is:

`\boxed{ x = -1}`