# How do you solve 3^x + 4 = 7^x - 1 ?

Oct 8, 2016

$x \approx 1.08777$

#### Explanation:

Calling

$f \left(x\right) = {7}^{x} - {3}^{x} - 5$ Expanding in Taylor series around ${x}_{0}$

$f \left(x\right) \approx f \left({x}_{0}\right) + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{{x}_{0}} \left(x - {x}_{0}\right)$

or

$f \left({x}_{1}\right) \approx f \left({x}_{0}\right) + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{{x}_{0}} \left({x}_{1} - {x}_{0}\right)$

If ${x}_{0}$ is near a function zero, then $f \left({x}_{1}\right) \approx 0$ so

${x}_{k + 1} = {x}_{k} - f \frac{{x}_{k}}{\frac{\mathrm{df}}{\mathrm{dx}}} _ \left({x}_{k}\right)$

Here

${\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{{x}_{k}} = {7}^{{x}_{k}} L o g \left(7\right) - {3}^{{x}_{k}} L o g \left(3\right)$

Begining with ${x}_{0} = 1$ we have

${x}_{0} = 1$
${x}_{1} = 1.09685$
${x}_{2} = 1.08786$
${x}_{3} = 1.08777$
${x}_{4} = 1.08777$