Solving for k in $ \log_{5} x = k \cdot \log_{10} x$

Question

If $ \log_{5} x = k \cdot \log_{10} x$

find the value of $k$ (rough approximation) without using calculator.

what i did is:

$ \log_{5} x = \log_{5}10 \cdot \log_{10}x$

$\therefore \:k = \log_{5}10 $

$\Rightarrow\: (1+4)^k = 10 $

$ 10 = (1+4)^k= 1+ 4k + \frac{k(k-1)}{ 2!}k^2 + \frac{k(k-1)(k-2)}{3!}k^3\cdots$

but after that i could not find a way to get the value of k.

[would be appreciated if someone solves the above series for k.]

I think you meant to write, $$\therefore k=\log_510,$$ which gives you what you want immediately, namely $1+\log_52.$ — Allawonder, Oct 20 '19 at 17:48

J.G. · Answer 1 · 2019-10-20T16:23:54.467

1

$$\frac{\ln x}{\ln 5}=\frac{k\ln x}{\ln 10}\implies k=\frac{\ln 10}{\ln 5}=1+\frac{\ln 2}{\ln 5}\approx 1.43$$ because $\frac{\ln 2}{\ln 5}\approx\frac37$, because $2^7\approx 5^3$.

edited Oct 20 '19 at 16:23

answered Oct 20 '19 at 16:18

J.G.

115,835

without using calculator :) – sakib Oct 20 '19 at 16:19
Alternatively, $\frac{\log 2 + \log 5}{\log 5} = 1 + \frac{\log 2}{\log 5} = 1 + \log_5 2$. – Travis Willse Oct 20 '19 at 16:22
@sakib Thanks; fixed. – J.G. Oct 20 '19 at 16:24
@TravisWillse OP already had the answer. He only made an oversight. – Allawonder Oct 20 '19 at 17:50
@allawonder I'm not sure what you're referring to, though it looks like some comments have been removed in the meantime. I was only pointing out an alternative expression for the answer. – Travis Willse Oct 20 '19 at 19:15

Solving for k in $ \log_{5} x = k \cdot \log_{10} x$

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