What is the answer to (4^{x+2})/(3^{x+1)}=16 ?

The answer to (4^{x+2})/(3^{x+1)}=16 is x= 1/(2log_{3)(2)-1}

Solve for x: (4^{x+2})/(3^{x+1)}=16

The solution is x= 1/(2log_{3)(2)-1}

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\frac{4 ^{x + 2}}{3 ^{x + 1}} = 16

x = \frac{1}{2 lo g _{3} ( 2 ) - 1}

Decimal Notation

x = 3.81884 \dots

Solution steps

\frac{4 ^{x + 2}}{3 ^{x + 1}} = 16

Apply exponent rules

(lo g_{3} (4) (x + 2) - x - 1) ln (3) = ln (16)

Solve

(lo g_{3} (4) (x + 2) - x - 1) ln (3) = ln (16) : x = \frac{1}{2 lo g _{3} ( 2 ) - 1}

x = \frac{1}{2 lo g _{3} ( 2 ) - 1}

Verify Solutions:

x = \frac{1}{2 lo g _{3} ( 2 ) - 1}

True

The solution is

x = \frac{1}{2 lo g _{3} ( 2 ) - 1}

2^{x} - 1 + 2^{x} + 1 = 2

7^{x - 1} + 7^{x} = 2744

\frac{1}{8} = \frac{4}{2 ^{\frac{t}{7}}}

5^{3 x - 1} = 25

3^{x} = \frac{14}{25}

What is the answer to (4^{x+2})/(3^{x+1)}=16 ?
The answer to (4^{x+2})/(3^{x+1)}=16 is x= 1/(2log_{3)(2)-1}
Solve for x: (4^{x+2})/(3^{x+1)}=16
The solution is x= 1/(2log_{3)(2)-1}