What is the answer to 3+2^x+1=[24-(2^2-1)] ?

The answer to 3+2^x+1=[24-(2^2-1)] is x=(ln(17))/(ln(2))

Solve for x: 3+2^x+1=[24-(2^2-1)]

The solution is x=(ln(17))/(ln(2))

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3 + 2^{x} + 1 = [24 - (2^{2} - 1)]

x = \frac{ln ( 17 )}{ln ( 2 )}

Decimal Notation

x = 4.08746 \dots

Solution steps

3 + 2^{x} + 1 = [24 - (2^{2} - 1)]

Group like terms

2^{x} + 3 + 1 = [24 - (2^{2} - 1)]

Add the numbers:

3 + 1 = 4

2^{x} + 4 = 24 - (2^{2} - 1)

Move

4

to the right side

2^{x} = - 2^{2} + 21

Simplify

2^{x} = 17

Apply exponent rules

x ln (2) = ln (17)

Solve

x ln (2) = ln (17) : x = \frac{ln ( 17 )}{ln ( 2 )}

x = \frac{ln ( 17 )}{ln ( 2 )}

e^{2} x - e^{x} - 6 = 0

3^{x} = 3 x + 2

x, 1 6^{2 x - 3} 4^{- 2 x} = 16

e^{- 4 x} = e

e^{30 k} = \frac{49953}{50000}

What is the answer to 3+2^x+1=[24-(2^2-1)] ?
The answer to 3+2^x+1=[24-(2^2-1)] is x=(ln(17))/(ln(2))
Solve for x: 3+2^x+1=[24-(2^2-1)]
The solution is x=(ln(17))/(ln(2))