What is 3^{4x-7}<4^{2x+3} ?

The solution to 3^{4x-7}<4^{2x+3} is x<(6ln(2)+7ln(3))/(4(-ln(2)+ln(3)))

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3^{4x-7}<4^{2x+3}

3^{4 x - 7} < 4^{2 x + 3}

x < \frac{6 ln ( 2 ) + 7 ln ( 3 )}{4 ( - ln ( 2 ) + ln ( 3 ) )}

Interval Notation

(- \infty, \frac{6 ln ( 2 ) + 7 ln ( 3 )}{4 ( - ln ( 2 ) + ln ( 3 ) )})

Decimal Notation

x < 7.30591 \dots

Solution steps

3^{4 x - 7} < 4^{2 x + 3}

f (x) < g (x)

then

ln (f (x)) < ln (g (x))

ln (3^{4 x - 7}) < ln (4^{2 x + 3})

Simplify

ln (3^{4 x - 7}) : ln (3) (4 x - 7)

Simplify

ln (4^{2 x + 3}) : 2 ln (2) (2 x + 3)

ln (3) (4 x - 7) < 2 ln (2) (2 x + 3)

ln (3) (4 x - 7) < 2 ln (2) (2 x + 3) : x < \frac{6 ln ( 2 ) + 7 ln ( 3 )}{4 ( - ln ( 2 ) + ln ( 3 ) )}

x < \frac{6 ln ( 2 ) + 7 ln ( 3 )}{4 ( - ln ( 2 ) + ln ( 3 ) )}

- 10 - 5

34 3^{\frac{4}{3}}

- 8 \geq - 5 x + 2 > - 38

3 x^{2} - x = 4

8^{- 2}

What is 3^{4x-7}<4^{2x+3} ?
The solution to 3^{4x-7}<4^{2x+3} is x<(6ln(2)+7ln(3))/(4(-ln(2)+ln(3)))