I need to solve $\log$ of $36$ in base $2$
The logarithm result $= x$.
$$ \log_ 2 36 = x. $$ How do I determine value of $x$ in
$$ 2^x=36 $$
I don't know how do it, since there's perfect square of this number.
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1You need to "solve" $\log_2 (36) = x$ – The Chaz 2.0 Oct 09 '13 at 04:47
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Or $26$ or whatever. – The Chaz 2.0 Oct 09 '13 at 04:47
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edited. Thanks. I'm a bit tired.. sorry for the mistakes – Assembly Oct 09 '13 at 04:48
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After reading this comment, you will have surely heard of the "change of base" formula for logarithms. You should use that if you would like to compute a decimal approximation using a hand calculator. – The Chaz 2.0 Oct 09 '13 at 04:50
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If the fact that it isn't an integer is tough for you, you can think of it as solving $2^n = 36^m$, then $x = n/m$. But if you are using an electronic calculator, then just find $log(36)/log(2) = log_2(36)$ – DanielV Oct 09 '13 at 05:09
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With a calculator, you can simply calculate: $$x= \log_2 36 = \log 36 / \log 2$$ Without a calculator, you know that $x$ must be a little over $5$, since $2^5=32$. Now: $$2^x = 2^{x-5}2^5 = 36 \to 2^{x-5}=36/32=1+1/8$$ Using the fact that for small $x$ $\log_b (1+x)\approx x/ \ln b$: $$(x-5)=\log_2(1+1/8)\approx \frac{1}{8\ln 2}$$ $$x\approx 5+\frac{1}{8\ln 2}\approx5+\frac{1}{5.6}$$
Nathaniel Bubis
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I'm going to solve
log 5, shouldn't it be something like10^x = 5? why do others solve it asx=10^5? it give very different result from my book and my calculator. – Assembly Oct 09 '13 at 06:50