How do you solve $3^{x} + 4 = 7^{x} - 1$ $3^x + 4 = 7^x - 1$ ?

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Answer 1 · 2016-10-08T20:01:49

$x \approx 1.08777$ $x approx 1.08777$

Calling

$f (x) = 7^{x} - 3^{x} - 5$ $f(x) = 7^x-3^x-5$ Expanding in Taylor series around $x_{0}$ $x_0$

$f (x) \approx f (x_{0}) + {(\frac{d f}{d x})}_{x_{0}} (x - x_{0})$ $f(x) approx f(x_0)+((df)/(dx))_(x_0)(x-x_0)$

or

$f (x_{1}) \approx f (x_{0}) + {(\frac{d f}{d x})}_{x_{0}} (x_{1} - x_{0})$ $f(x_1) approx f(x_0)+((df)/(dx))_(x_0)(x_1-x_0)$

If $x_{0}$ $x_0$ is near a function zero, then $f (x_{1}) \approx 0$ $f(x_1) approx 0$ so

$x_{k + 1} = x_{k} - \frac{f (x_{k})}{{(\frac{d f}{d x})}_{x_{k}}}$ $x_(k+1)=x_k-f(x_k)/((df)/(dx))_(x_k)$

Here

${(\frac{d f}{d x})}_{x_{k}} = 7_{k}^{x_{k}} log (7) - 3_{k}^{x_{k}} log (3)$ $((df)/(dx))_(x_k)=7^(x_k) Log(7)-3^(x_k) Log(3)$

Begining with $x_{0} = 1$ $x_0 = 1$ we have

$x_{0} = 1$ $x_0=1$
$x_{1} = 1.09685$ $x_1=1.09685$
$x_{2} = 1.08786$ $x_2=1.08786$
$x_{3} = 1.08777$ $x_3=1.08777$
$x_{4} = 1.08777$ $x_4=1.08777$

How do you solve 3^x + 4 = 7^x - 1 3x+4=7x−13^x + 4 = 7^x - 1 ?