What is the answer to log_{7}(2^{4+3x})=10 ?

The answer to log_{7}(2^{4+3x})=10 is x=(ln(282475249)-4ln(2))/(3ln(2))

Solve for x: log_{7}(2^{4+3x})=10

The solution is x=(ln(282475249)-4ln(2))/(3ln(2))

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lo g_{7} (2^{4 + 3 x}) = 10

x = \frac{ln ( 282475249 ) - 4 ln ( 2 )}{3 ln ( 2 )}

Solution steps

lo g_{7} (2^{4 + 3 x}) = 10

Rewrite the equation with

2^{4 + 3 x} = u

lo g_{7} (u) = 10

Solve

lo g_{7} (u) = 10 : u = 282475249

u = 282475249

Substitute back

u = 2^{4 + 3 x},

solve for

x

Solve

2^{4 + 3 x} = 282475249 : x = \frac{ln ( 282475249 ) - 4 ln ( 2 )}{3 ln ( 2 )}

x = \frac{ln ( 282475249 ) - 4 ln ( 2 )}{3 ln ( 2 )}

- 5 \cdot e^{10 t} = - 30

y, x^{y} = 3

e^{- x} = 4

e^{- x} = 6

What is the answer to log_{7}(2^{4+3x})=10 ?
The answer to log_{7}(2^{4+3x})=10 is x=(ln(282475249)-4ln(2))/(3ln(2))
Solve for x: log_{7}(2^{4+3x})=10
The solution is x=(ln(282475249)-4ln(2))/(3ln(2))