logarithms and exponents with base 10

Calculus Level 3

Compute 1 0 x x log ⁡ 10 ( 1 0 x + x ) − 1 0 x \frac{10^x}{x} \log_{10}(10^x + x) - 10^x , where x = 1 0 10 x = 10^{10} .

Inspired by this problem.


The answer is 0.4343.

Samuel Lo by Samuel Lo , 29, Hong Kong

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Samuel Lo
Sep 19, 2018

1 0 x x log ⁡ 10 ( 1 0 x + x ) − 1 0 x = 1 0 x x log ⁡ 10 [ 1 0 x ( 1 + x 1 0 − x ) ] − 1 0 x = 1 0 x x [ log ⁡ 10 ( 1 0 x ) + log ⁡ 10 ( 1 + x 1 0 − x ) ] − 1 0 x = 1 0 x x [ x + log ⁡ 10 ( 1 + x 1 0 − x ) ] − 1 0 x = 1 0 x + 1 0 x x log ⁡ 10 ( 1 + x 1 0 − x ) − 1 0 x = 1 0 x x log ⁡ 10 ( 1 + x 1 0 − x ) = 1 0 x x ln ⁡ ( 1 + x 1 0 − x ) ln ⁡ 10 = 1 0 x x ln ⁡ 10 ln ⁡ ( 1 + x 1 0 − x ) = 1 0 x x ln ⁡ 10 [ x 1 0 − x − ( x 1 0 − x ) 2 2 + … ] ≈ 1 0 x x ln ⁡ 10 x 1 0 − x = 1 ln ⁡ 10 ≈ 0.4343 \begin{aligned} & \frac{10^x}{x} \log_{10}(10^x + x) - 10^x \\ =& \frac{10^x}{x} \log_{10}[10^x (1 + x 10^{-x})] - 10^x \\ =& \frac{10^x}{x} [\log_{10}(10^x) + \log_{10}(1 + x 10^{-x})] - 10^x \\ =& \frac{10^x}{x} [ x + \log_{10}(1 + x 10^{-x})] - 10^x \\ =& 10^x + \frac{10^x}{x} \log_{10}(1 + x 10^{-x}) - 10^x \\ =& \frac{10^x}{x} \log_{10}(1 + x 10^{-x}) \\ =& \frac{10^x}{x} \frac { \ln (1 + x 10^{-x}) }{\ln 10} \\ =& \frac{10^x}{x \ln 10} \ln (1 + x 10^{-x}) \\ = & \frac{10^x}{x \ln 10}[ x 10^{-x} - \frac{ (x 10^{-x})^2 }{2} + \dots ] \\ \approx & \frac{10^x}{x \ln 10} x 10^{-x} \\ =& \frac{1}{\ln 10} \\ \approx & 0.4343 \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...