What is the answer to (3^{x-1})(5^{2x+1})=15 ?

The answer to (3^{x-1})(5^{2x+1})=15 is x= 2/(1+2log_{3)(5)}

Solve for x: (3^{x-1})(5^{2x+1})=15

The solution is x= 2/(1+2log_{3)(5)}

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(3^{x - 1}) (5^{2 x + 1}) = 15

x = \frac{2}{1 + 2 lo g _{3} ( 5 )}

Decimal Notation

x = 0.50891 \dots

Solution steps

(3^{x - 1}) (5^{2 x + 1}) = 15

Apply exponent rules

(x - 1 + lo g_{3} (5) (2 x + 1)) ln (3) = ln (15)

Solve

(x - 1 + lo g_{3} (5) (2 x + 1)) ln (3) = ln (15) : x = \frac{2}{1 + 2 lo g _{3} ( 5 )}

x = \frac{2}{1 + 2 lo g _{3} ( 5 )}

(\frac{1}{8})^{x} = 2

9^{2 x} = 9^{0}

9^{2 x} = 9^{8}

x^{x} = 8

x^{x} = 7

What is the answer to (3^{x-1})(5^{2x+1})=15 ?
The answer to (3^{x-1})(5^{2x+1})=15 is x= 2/(1+2log_{3)(5)}
Solve for x: (3^{x-1})(5^{2x+1})=15
The solution is x= 2/(1+2log_{3)(5)}