What is the critical 3^x-2^x ?

The critical 3^x-2^x is x=(ln(ln(2))-ln(ln(3)))/(ln(3)-ln(2))

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critical 3^x-2^x

critical points

3^{x} - 2^{x}

x = \frac{ln ( ln ( 2 ) ) - ln ( ln ( 3 ) )}{ln ( 3 ) - ln ( 2 )}

Solution steps

3^{x} - 2^{x}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{ln ( ln ( 2 ) ) - ln ( ln ( 3 ) )}{ln ( 3 ) - ln ( 2 )}

Identify critical points not in

f (x)

domain

Domain of

3^{x} - 2^{x} : - \infty < x < \infty

All critical points are in domain

x = \frac{ln ( ln ( 2 ) ) - ln ( ln ( 3 ) )}{ln ( 3 ) - ln ( 2 )}

x^{2} - 8 x + 7

- x + 4

f (x) = - 9 x - 8 + 5

f (x) = lo g_{\frac{1}{2}} (x)

f (x) = x^{2} - x + 1

What is the critical 3^x-2^x ?
The critical 3^x-2^x is x=(ln(ln(2))-ln(ln(3)))/(ln(3)-ln(2))