How to proceed with logarithms in exponents like in this problem

Question

The product of all the solutions of the equation $x^{1+\log_{10}x} = 100000x$ is $$(A)~ 10 \qquad (B)~ 10^5 \qquad (C)~ 10^{-5} \qquad (D)~1$$

Is there some properties I should know to solve this?

Answer 1

If $u$ is a solution then $$\left(\frac1u\right)^{1+\log_{10}(1/u)}=\left(\frac1u\right)^{1-\log_{10}u}=u^{-1+\log_{10}u}=\frac{10000}u$$ We see that if $u$ is a solution, then so is $1/u$. Hence, the product of the solutions is $1$.

Answer 2

Hint: Using the below properties, you should be able to take the $\log_{10}$ of both sides and simplify, such that you can solve for $\log_{10} x$.

• $\log_b(x^a) = a\log_b(x)$
• $\log_b(a\cdot c) = \log_b(a) \cdot \log_b(c)$
• $x^{a+b} = x^ax^b$