What is the answer to 4^x+48=19*2^x ?

The answer to 4^x+48=19*2^x is x=4,x=(ln(3))/(ln(2))

Solve for x: 4^x+48=19*2^x

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

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4^{x} + 48 = 19 \cdot 2^{x}

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

Decimal Notation

x = 4, x = 1.58496 \dots

Solution steps

4^{x} + 48 = 19 \cdot 2^{x}

Apply exponent rules

(2^{x})^{2} + 48 = 19 \cdot 2^{x}

Rewrite the equation with

2^{x} = u

(u)^{2} + 48 = 19 u

Solve

u^{2} + 48 = 19 u : u = 16, u = 3

u = 16, u = 3

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

Solve

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

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

64 (4^{2 x}) - 20 (4^{x}) + 1 = 0

3 \cdot 9^{x} - 10 \cdot 3^{x} + 3 = 0

6^{x} = 23

x, 0 = e^{x y}

1 = 4^{- x}

What is the answer to 4^x+48=19*2^x ?
The answer to 4^x+48=19*2^x is x=4,x=(ln(3))/(ln(2))
Solve for x: 4^x+48=19*2^x
The solution is x=4,x=(ln(3))/(ln(2))