What is the answer to 10^x+10^{-x}=4 ?

The answer to 10^x+10^{-x}=4 is x=(ln(2+sqrt(3)))/(ln(10)),x=(ln(2-sqrt(3)))/(ln(10))

Solve for x: 10^x+10^{-x}=4

The solution is x=(ln(2+sqrt(3)))/(ln(10)),x=(ln(2-sqrt(3)))/(ln(10))

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1 0^{x} + 1 0^{- x} = 4

x = \frac{ln ( 2 + 3 )}{ln ( 10 )}, x = \frac{ln ( 2 - 3 )}{ln ( 10 )}

Decimal Notation

x = 0.57194 \dots, x = - 0.57194 \dots

Solution steps

1 0^{x} + 1 0^{- x} = 4

Apply exponent rules

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

Rewrite the equation with

1 0^{x} = u

u + (u)^{- 1} = 4

Solve

u + u^{- 1} = 4 : u = 2 + 3, u = 2 - 3

u = 2 + 3, u = 2 - 3

Substitute back

u = 1 0^{x},

solve for

x

Solve

1 0^{x} = 2 + 3 : x = \frac{ln ( 2 + 3 )}{ln ( 10 )}

Solve

1 0^{x} = 2 - 3 : x = \frac{ln ( 2 - 3 )}{ln ( 10 )}

x = \frac{ln ( 2 + 3 )}{ln ( 10 )}, x = \frac{ln ( 2 - 3 )}{ln ( 10 )}

(d^{2} - 9) = (e^{3})^{x} + e^{- 3} x

5^{x} (5^{x} - 4) + 3 = 0

(- 15)^{x} = 1

1 0^{2 x - 3} = (0.1)^{2}

x, 4^{x + y} = 3^{x^{2} + y^{2}}

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