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

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

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

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

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

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

Decimal Notation

x = - 2.70951 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

4^{(2 x - 1)} = \frac{1}{2}

2^{x + 4} = 5

2^{3 x} = 6^{x + 2}

5^{2 x - 13} = 25

\frac{2 e ^{x}}{4} = 9

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