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?
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kccu
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user642771
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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 $1$.
ajotatxe
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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 $\log_{10} x$.
- $\log_b(x^a) = a\log_b(x)$
- $\log_b(a\cdot c) = \log_b(a) \cdot \log_b(c)$
- $x^{a+b} = x^ax^b$
Dando18
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$$x^{1+\log_{10} x} = x\cdot x^{\log_{10} x} = 100000x$$
This leads to $$x^{\log_{10} x} = 100000$$
– JacobCheverie Apr 24 '19 at 13:03