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How do you solve $$1.08=(1+x)^{1200}?$$ Thanks so much for any guidance.

Iuli
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nishman
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5 Answers5

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Computing an approximation by hand. We have $$ x=(1+0.08)^{1/1200}-1. $$ Using the Taylor expansion $(1+x)^a=1+a\,x+O(x^2)$ we can approximate the answer as $$ x\approx\frac{0.08}{1200}=\frac{1}{15000}. $$

Julián Aguirre
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You have

$$\log(1.08)=\log\left((1+x)^{1200}\right)=1200\log(1+x)$$

so

$$\log(1+x)=\frac{\log(1.08)}{1200}$$

finally,

$$x=\exp\left(\frac{\log(1.08)}{1200}\right)-1\approx 6.41\cdot 10^{-5}.$$

E. Joseph
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You're (all) missing the second solution

$$x=\color{red}{-}\exp\left(\frac{\log(1.08)}{1200}\right)-1,$$ which can be equivalently written as $$x=\color{red}{-}\sqrt[1200]{1.08}-1.$$

b00n heT
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  • and 1198 other complex solutions with non-zero imaginary part... – Claude Oct 22 '16 at 16:29
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    Although you are most definitely correct, the fact that the exponent is even does play a role qhen considering real solutions only, so I wanted to point that out. – b00n heT Oct 22 '16 at 16:46
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I believe you are interested in method how to find the decimals?

So you have $$x=exp(\frac{log 1.08}{1200})-1$$ You can use Taylor series(google it...) $$Log1.08=0+0.08-0.0032+...~=0.077$$ Divide by 1200... ~= 0.000064 And so on...

Djura Marinkov
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$1.08=(1+x)^{1200}$ implies clearly that $x$ must be small, very small indeed but not negative as I believed at the first sight because $a^x\gt 1$ implies $x\ln a\gt 0$ which is not the case. Calculating, as usual with these kind of numbers, one has $$(1+10^{-4})^{1200}\approx 1.1274900870\\(1+10^{-5})^{1200}\approx 1.0120722814$$ hence $$10^{-5}\lt x\lt 10^{-4}$$ Paying more attention to these calculations, (or, for example, expanding in series) we can get the approximation $\color{red}{x=6.44\cdot 10^{-5}}$ which gives $$(1+6.44\cdot10^{-5})^{1200}\approx 1.0803418422$$

Piquito
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