What is the limit as x approaches infinity+of (log_{4}(x))/(log_{7)(x)} ?

The limit as x approaches infinity+of (log_{4}(x))/(log_{7)(x)} is (ln(7))/(2ln(2))

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limit as x approaches infinity+of (log_{4}(x))/(log_{7)(x)}

x \to \infty + lim (\frac{lo g _{4} ( x )}{lo g _{7} ( x )})

\frac{ln ( 7 )}{2 ln ( 2 )}

Decimal

1.40367 \dots

Solution steps

x \to \infty lim (\frac{lo g _{4} ( x )}{lo g _{7} ( x )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{\frac{1}{2 l n ( 2 ) x}}{\frac{1}{x l n ( 7 )}})

\frac{\frac{1}{2 l n ( 2 ) x}}{\frac{1}{x l n ( 7 )}} = \frac{ln ( 7 )}{2 ln ( 2 )}

= x \to \infty lim (\frac{ln ( 7 )}{2 ln ( 2 )})

x \to a lim c = c

= \frac{ln ( 7 )}{2 ln ( 2 )}

\frac{d}{d x} (\frac{x ^{2}}{x ^{3} + 4})

\frac{d y}{d x} = y^{3} x e^{x^{2}}

y^{^{''}} - 8 y = 0

a x^{^{''}} + b x = 0

\frac{d}{d x} (e^{t a n h (2 x)})

What is the limit as x approaches infinity+of (log_{4}(x))/(log_{7)(x)} ?
The limit as x approaches infinity+of (log_{4}(x))/(log_{7)(x)} is (ln(7))/(2ln(2))