What is solution of this logarithmic equation

Question

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.

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 06 '17 at 08:15

score 2 · Accepted Answer · 2017-07-07T09:29:15.817

2

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?

edited Jul 07 '17 at 09:29

answered Jul 06 '17 at 08:18

In your first line, it should be $ 5^u - 3^{u-1} $ or not? Please specify – Fghj Jul 06 '17 at 08:23
Please note that I've edited my question. Is your answer still the same? – Fghj Jul 06 '17 at 08:26
@RaviPrakash If the original algebra has been changed then I will have to go back and redo. Hold on... – Jul 06 '17 at 08:29
Yes. I regret for that. – Fghj Jul 06 '17 at 08:32
@RaviPrakash It is now edited. Thanks – Jul 06 '17 at 08:33
But please tell me, how you wrote $ 6 × 5^{u-1} = 10 × 3^{u-1} $ in the third line? – Fghj Jul 07 '17 at 09:26
Ok edited to reflect this step. – Jul 07 '17 at 09:29

What is solution of this logarithmic equation

1 Answers1