`3(5^(x-1)) = 86` Solve the equation accurate to three decimal places

Tuesday, December 29, 2009

Problem: $\displaystyle{3}{\left({{5}}^{{{x}-{1}}}\right)}={86}$

To simplify, we divide both sides by 3:

$\displaystyle\frac{{{3}{\left({{5}}^{{{x}-{1}}}\right)}}}{{3}}=\frac{{{86}}}{{3}}$

$\displaystyle{{5}}^{{{x}-{1}}}=\frac{{{86}}}{{3}}$

Take the "log" on both sides to apply the logarithm property: $\displaystyle{\log{{\left({{x}}^{{y}}\right)}}}={y}\cdot{\log{{\left({x}\right)}}}$ .

Applying on the given problem:

$\displaystyle{\log{{\left({{5}}^{{{x}-{1}}}\right)}}}={\log{{\left(\frac{{{86}}}{{3}}\right)}}}$

$\displaystyle{\left({x}-{1}\right)}{\log{{\left({5}\right)}}}={\log{{\left(\frac{{{86}}}{{3}}\right)}}}$

Divide both sides by $\displaystyle{\log{{\left({5}\right)}}}$ to isolate $\displaystyle{\left({x}-{1}\right)}$ :

Add 1 on both sides to solve x:

$\displaystyle{x}-{1}=\frac{{{\log{{\left(\frac{{{86}}}{{3}}\right)}}}}}{{{\log{{\left({5}\right)}}}}}$

+1 +1

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$\displaystyle{x}=\frac{{{\log{{\left(\frac{{{86}}}{{3}}\right)}}}}}{{{\log{{\left({5}\right)}}}}}+{1}$

$\displaystyle{x}\approx{3.085}$

To check, plug-in $\displaystyle{x}={3.085}$ in :

$\displaystyle{3}{\left({{5}}^{{{3.085}-{1}}}\right)}=?{86}$

$\displaystyle{3}\cdot{{5}}^{{2.085}}=?{86}$

$\displaystyle{3}\cdot{28.66503386}=?{86}$

$\displaystyle{85.99510157}\approx{86}$ TRUE.

Conclusion: x= 3.085 as the real solution

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