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I understand that

$\log_b n = x \iff b^x = n$

But all examples I see is with values that I naturally know how to calculate (like $2^x = 8, x=3$)

What if I don't? For example, how do I solve for $x$ when:

$$\log_{1.03} 2 = x\quad ?$$

$$\log_{8} 33 = x\quad ?$$

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    $\begingroup$ Really, the only way to go about solving logarithms in most cases is by using a calculator. You could solve a rounded answer by hand using a Taylor series expansion, but that would be exhausting. $\endgroup$
    – Cisplatin
    Commented Jun 9, 2013 at 14:57
  • $\begingroup$ You can get rid of the weird logarithm base by using your first equation gong left to right; then go right to left using logarithms you can access on your calculator $\endgroup$
    – DJohnM
    Commented Jun 9, 2013 at 14:58
  • $\begingroup$ Fabricio: How does the accepted answer allows you "to find the value without a calculator"? $\endgroup$
    – Did
    Commented Jun 9, 2013 at 16:09

5 Answers 5

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The logarithm $\log_{b} (x)$ can be computed from the logarithms of $x$ and $b$ with respect to a positive base $k$ using the following formula:

$$\log_{b} (x) = \frac{\log_{k} (x)}{\log_{k} (b)}.$$

So your examples can be solved in the following way with a calculator:

$$x = \log_{1.03} (2) = \frac{\log_{10} (2)}{\log_{10} (1.03)} = \frac{0.301}{0.013} = 23.450, $$

$$x = \log_{8} (33) = \frac{\log_{10} (33)}{\log_{10} (8)} = \frac{1.519}{0.903} = 1.681.$$

If you know that $b$ and $x$ are both powers of some $k$, then you can evaluate the logarithm without a calculator by the power identity of logarithms, e.g.,

$$x = \log_{81} (27) = \frac{\log_{3} (27)}{\log_{3} (81)} = \frac{\log_{3} (3^3)}{\log_{3} (3^4)} = \frac{3 \cdot \log_{3} (3)}{4 \cdot \log_{3} (3)} = \frac{3}{4}.$$

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method 1: use a calculator

method 2 (more fun): $\log_b a=\frac{\ln b}{\ln a}$

To calculate natural logs, if $|x|<1$ use the power series $\ln (x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$ and if not find the log of the reciprocal and subtract from zero. Powers of $x$ can be calculated by convolving as power series in $10$.

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  • $\begingroup$ I'm very 'noob' with all this. Can you provide a link where I can understand all these matters? $\endgroup$
    – Fabricio
    Commented Jun 9, 2013 at 15:09
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There's actually nothing to solve here. You have an exact expression for $x$ in each case- you may not know exactly what number it corresponds to, just like you might not know what the square root of three equals when you solve something like $x^2=3$ but it's still just a number- in the first case, it would be the power to which I raise 1.03 to get 2 which is approximately 23.45

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Starting with :$$\log_{1.03} 2 = x$$This is exactly the same as $$1.03^x=2$$ now take logarithms of both sides$$x \times \log(1.03)=\log(2)$$ Now, use your calculator

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  • $\begingroup$ Isn't there how to calculate without the calculator? I want to find the value without a calculator. $\endgroup$
    – Fabricio
    Commented Jun 9, 2013 at 15:06
  • $\begingroup$ @AmWhy: I wanted to show where the substitution you used (the correct one, of course) came from... $\endgroup$
    – DJohnM
    Commented Jun 9, 2013 at 15:20
  • $\begingroup$ Good point! ;-) $\endgroup$
    – amWhy
    Commented Jun 9, 2013 at 15:21
  • $\begingroup$ I just touched up the formatting in your answer, by adding two backslashes. Hope you don't mind! $\endgroup$
    – amWhy
    Commented Jun 9, 2013 at 15:26
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$$\log_bn=\frac{\ln n}{\ln b}=\frac{\log_{10}n}{\log_{10}b}=\frac{\log_2n}{\log_2b}=\ldots$$ Edit: Now added in a comment:

I want to find the value without a calculator.

...Which is an entirely new take on the question.

To get $x=\log_{1.03}2$, one could compute the successive powers of $1.03$. If $1.03^n\lt2\lt1.03^{n+1}$, then $n\lt x\lt n+1$. The rest depends on your ability to compute $1.03^n$ for $n$ in the $20$-$30$ range... but this yields $n=23$.

Another approach is to use power series, in this case $x=y/z$ with $y=\ln2$ and $z=\ln1.03$. Using the expansion $\log1+t=\sum\limits_{n\geqslant1}(-1)^{n-1}\frac{t^n}n$, one gets $$ y=-\ln\left(1-\tfrac12\right)=\sum_{n\geqslant1}\frac1{n2^n},\qquad z=\ln(1+.03)=\sum\limits_{n\geqslant1}(-1)^{n-1}\frac{(.03)^n}n. $$ Keeping $9$ terms in $y$ and $2$ terms in $z$ yields an error of at most $10^{-4}$.

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