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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 $10$.

What I had done:

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

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.

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    $\begingroup$ 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. $\endgroup$ – Osama Ghani Jul 6 '17 at 8:15
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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}

Can you take it from here?

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  • $\begingroup$ In your first line, it should be $ 5^u - 3^{u-1} $ or not? Please specify $\endgroup$ – Ravi Prakash Jul 6 '17 at 8:23
  • $\begingroup$ Please note that I've edited my question. Is your answer still the same? $\endgroup$ – Ravi Prakash Jul 6 '17 at 8:26
  • $\begingroup$ @RaviPrakash If the original algebra has been changed then I will have to go back and redo. Hold on... $\endgroup$ – complexmanifold Jul 6 '17 at 8:29
  • $\begingroup$ Yes. I regret for that. $\endgroup$ – Ravi Prakash Jul 6 '17 at 8:32
  • $\begingroup$ @RaviPrakash It is now edited. Thanks $\endgroup$ – complexmanifold Jul 6 '17 at 8:33

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