What is the answer to 2^{2x+5}0.6^x=12^x ?

The answer to 2^{2x+5}0.6^x=12^x is x= 5/(log_{2)(20)-2}

Solve for x: 2^{2x+5}0.6^x=12^x

The solution is x= 5/(log_{2)(20)-2}

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2^{2 x + 5} 0. 6^{x} = 1 2^{x}

x = \frac{5}{lo g _{2} ( 20 ) - 2}

Decimal Notation

x = 2.15338 \dots

Solution steps

2^{2 x + 5} \cdot 0. 6^{x} = 1 2^{x}

Apply exponent rules

2 x + 5 + lo g_{2} (\frac{3}{5}) x = lo g_{2} (12) x

Solve

2 x + 5 + lo g_{2} (\frac{3}{5}) x = lo g_{2} (12) x : x = \frac{5}{lo g _{2} ( 20 ) - 2}

x = \frac{5}{lo g _{2} ( 20 ) - 2}

9^{x - 2} = 2 7^{x}

x^{4 x} = 2

2 (8^{x}) - 1 = 9

a \cdot b^{t} = c \cdot d^{t}

2^{1000} = 4^{x}

What is the answer to 2^{2x+5}0.6^x=12^x ?
The answer to 2^{2x+5}0.6^x=12^x is x= 5/(log_{2)(20)-2}
Solve for x: 2^{2x+5}0.6^x=12^x
The solution is x= 5/(log_{2)(20)-2}