# What is solution of this logarithmic equation

I am new to logarithms. I've tried to solve this but I couldn't. Below is the equation,

$$5^{\log x} - 3^{\log(x) -1} = 3^{\log(x) +1} - 5 ^{\log(x) -1}$$

Base of $\log$ is $10$.

$$5^{\log x} + 5^{\log(x) -1} = 3^{\log(x) +1} + 3^{\log(x) -1}$$

And tried taking $\log$ on both sides.

But I am stuck at the fact that what should be the result of something like $log (k^{\log x} + k^{\log(x) -1})$ , where $k$ is any constant , which is exactly the thing at LHS and $log (k^{\log(x) +1} + k^{\log(x) -1})$ RHS of my above equation.

• Note that the left is the same as $5^{log x}(1+5)= 6 \times 5^{log x}$.Try a similar simplification on the right. – Osama Ghani Jul 6 '17 at 8:15

Let $u = \log x$ then \begin{align} 5^u-3^{u-1} &= 3^{u+1}-5^{u-1} \\ \implies 5^{u}+5^{u-1} &= 3^{u+1}+3^{u-1} \\ \implies 5^{u-1}(5+1) &= 3^{u-1}(3^2+1)\\ \implies 6\cdot 5^{u-1} &= 10\cdot 3^{u-1} \\ \implies 3\cdot 5^{u-1} &= 5\cdot 3^{u-1} \\ \implies \log 3 + (u-1) \log 5 &= \log 5 + (u-1) \log 3 \\ &\vdots \end{align}
• In your first line, it should be $5^u - 3^{u-1}$ or not? Please specify – Ravi Prakash Jul 6 '17 at 8:23