Calculating integer solutions of a logarithmic equation

Question

The question asks:
Calculate the integer solutions of the equation $\log_2(x+2)+\frac12\log_2(x-5)^2=3$
To me, this is trivial if solved in the following way:
$(x+2)+(x-5)=2^3$
$2x-3=8$
Answer: $x=\dfrac{11}{2}=5.5$

However, the question asks to compute the integer solutions. Am I missing out something here? Thanks!

Answer 1

You just made a mistake that $$\color{red}{\log_d a + \log_d b =c}$$ $$\color{red}{a+b=d^c}$$ $$\color{green}{\log_d a + \log_d b=\log_d a b}$$ $$\log_2(x+2)+\frac{1}{2}\log_2(x-5)^2=3$$ $$\log_2(x+2)+\frac{2}{2}\log_2(x-5)=3$$ $$\log_2(x+2)+\log_2(x-5)=3$$ $$\log_2\ (x+2)(x-5)=3$$ $$x=6$$ $\color{red}{x\ne -3}$ because it is not in domain .

Answer 2

You should have $(x+2)(x-5)=2^3$. The exponent of a sum is the product of the exponents.

Answer 3

$log(a)+log(b)=log(a\cdot b)$

Solution: $log_2(x+2)(x-5)=3$ $\longrightarrow$ $(x+2)(x-5)=2^3$ $\longrightarrow$ $x^2-3x-10=8$ $\longrightarrow$ so the answer is x=6 and x=-3. I hope this helped.