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

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Answer 1 · 2016-10-08T20:01:49

$x approx 1.08777$

Calling

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

$f(x) approx f(x_0)+((df)/(dx))_(x_0)(x-x_0)$

or

$f(x_1) approx f(x_0)+((df)/(dx))_(x_0)(x_1-x_0)$

If $x_0$ is near a function zero, then $f(x_1) approx 0$ so

$x_(k+1)=x_k-f(x_k)/((df)/(dx))_(x_k)$

Here

$((df)/(dx))_(x_k)=7^(x_k) Log(7)-3^(x_k) Log(3)$

Begining with $x_0 = 1$ we have

$x_0=1$
$x_1=1.09685$
$x_2=1.08786$
$x_3=1.08777$
$x_4=1.08777$

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