How to solve the following equation without using a calculator?


I'm not getting a solution without using a machine calculation or logarithm table.


closed as unclear what you're asking by user147263, Daniel Fischer, PVAL-inactive, apnorton, M. Vinay Aug 6 '14 at 1:51

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  • 2
    $\begingroup$ Hint: How can you use the property $b \log a = \log(a^b)$? $\endgroup$ – Sammy Black Apr 14 '13 at 0:10
  • $\begingroup$ Can you provide more detail on what you are studying? Are you expected to use printed log tables? Are you expected to express the answer as a number or as an expression? $\endgroup$ – half-integer fan Apr 14 '13 at 0:13
  • $\begingroup$ $1.04$ is 'close' to $1$. $\endgroup$ – Felix Marin Aug 8 '14 at 19:21

I believe you are using a European convention where the comma is a decimal point, so I will assume this is the case.

Use the Maclurin series for log:

$$\log{(1+z)} = z - \frac{z^2}{2} + \frac{z^3}{3} - \ldots$$

You did not specify an accuracy, so we'll estimate the error from the quadratic approximation. You have $z=0.04$; then

$$\log{1.04} \approx 0.04 - \frac{0.04^2}{2} = 0.04 - 0.0008 = 0.0392$$

The error is about $0.04^3/3 = 0.000021\bar{3}$.

Then $\log{1.04}/6 \approx 0.0065\bar{3}$.

Now use the expansion

$$e^y \approx 1+y+\frac{y^2}{2} + \frac{y^3}{6} + \ldots$$

where now $y = 0.0065\bar{3}$


$$x \approx 1+0.0065\bar{3} + \frac{0.0065\bar{3}^2}{2} $$

$$0.065\bar{3}^2 \approx 0.000021$$

Then $x \approx 1.00655$.


$$log(x)=\dfrac{log(1,04)}{6} \Leftrightarrow 6 \log(x)=\log(1,04) \Leftrightarrow \log(x^6)=\log(1,04) \Leftrightarrow x= \sqrt[6]{1.04}$$

The question is, are you happy with this answer, or do you need the exact decimal expression?


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