What is 62^{x+3}<= 7^{2x+1} ?

The solution to 62^{x+3}<= 7^{2x+1} is x<= (ln(7)-3ln(62))/(ln(62)-2ln(7))

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62^{x+3}<= 7^{2x+1}

6 2^{x + 3} \leq 7^{2 x + 1}

x \leq \frac{ln ( 7 ) - 3 ln ( 62 )}{ln ( 62 ) - 2 ln ( 7 )}

Interval Notation

(- \infty, \frac{ln ( 7 ) - 3 ln ( 62 )}{ln ( 62 ) - 2 ln ( 7 )}]

Decimal Notation

x \leq - 44.34708 \dots

Solution steps

6 2^{x + 3} \leq 7^{2 x + 1}

f (x) \leq g (x)

then

ln (f (x)) \leq ln (g (x))

ln (6 2^{x + 3}) \leq ln (7^{2 x + 1})

Simplify

ln (6 2^{x + 3}) : ln (62) (x + 3)

Simplify

ln (7^{2 x + 1}) : ln (7) (2 x + 1)

ln (62) (x + 3) \leq ln (7) (2 x + 1)

ln (62) (x + 3) \leq ln (7) (2 x + 1) : x \leq \frac{ln ( 7 ) - 3 ln ( 62 )}{ln ( 62 ) - 2 ln ( 7 )}

x \leq \frac{ln ( 7 ) - 3 ln ( 62 )}{ln ( 62 ) - 2 ln ( 7 )}

16 z^{4} - 81

(3 x - 4)^{2} = 21

3 x - 1 < 9

y^{2} - \frac{1}{9} = 0

x^{2} - 3 x + 2 x y - 6 y

What is 62^{x+3}<= 7^{2x+1} ?
The solution to 62^{x+3}<= 7^{2x+1} is x<= (ln(7)-3ln(62))/(ln(62)-2ln(7))