What is the answer to 3^x-(sqrt(3))^{x+4}+20=0 ?

The answer to 3^x-(sqrt(3))^{x+4}+20=0 is x=(2ln(5))/(ln(3)),x=(4ln(2))/(ln(3))

Solve for x: 3^x-(sqrt(3))^{x+4}+20=0

The solution is x=(2ln(5))/(ln(3)),x=(4ln(2))/(ln(3))

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3^{x} - (3)^{x + 4} + 20 = 0

x = \frac{2 ln ( 5 )}{ln ( 3 )}, x = \frac{4 ln ( 2 )}{ln ( 3 )}

Decimal Notation

x = 2.92994 \dots, x = 2.52371 \dots

Solution steps

3^{x} - (3)^{x + 4} + 20 = 0

Apply exponent rules

3^{x} - 3^{x} \cdot 3^{2} + 20 = 0

Rewrite the equation with

3^{x} = u

u - u \cdot 3^{2} + 20 = 0

Solve

u - u \cdot 3^{2} + 20 = 0 : u = 25, u = 16

u = 25, u = 16

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 25 : x = \frac{2 ln ( 5 )}{ln ( 3 )}

Solve

3^{x} = 16 : x = \frac{4 ln ( 2 )}{ln ( 3 )}

x = \frac{2 ln ( 5 )}{ln ( 3 )}, x = \frac{4 ln ( 2 )}{ln ( 3 )}

1 8^{x} - 8 \cdot 6^{x} - 9 \cdot 2^{x} = 0

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

2^{x} \cdot 2^{4} = 2^{2 x}

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

5^{2 x^{2} - 3 x - 2} = 1

What is the answer to 3^x-(sqrt(3))^{x+4}+20=0 ?
The answer to 3^x-(sqrt(3))^{x+4}+20=0 is x=(2ln(5))/(ln(3)),x=(4ln(2))/(ln(3))
Solve for x: 3^x-(sqrt(3))^{x+4}+20=0
The solution is x=(2ln(5))/(ln(3)),x=(4ln(2))/(ln(3))