What is the answer to (a^x+1)/(a^x)=n ?

The answer to (a^x+1)/(a^x)=n is x=(ln(-1/(1-n)))/(ln(a))

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(a^x+1)/(a^x)=n

\frac{a ^{x} + 1}{a ^{x}} = n

x = \frac{ln ( - \frac{1}{1 - n} )}{ln ( a )}

Solution steps

\frac{a ^{x} + 1}{a ^{x}} = n

Apply exponent rules

(a^{x} + 1) (a^{x})^{- 1} = n

Rewrite the equation with

a^{x} = u

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

Solve

(u + 1) u^{- 1} = n : u = - \frac{1}{1 - n}

u = - \frac{1}{1 - n}

Substitute back

u = a^{x},

solve for

x

Solve

a^{x} = - \frac{1}{1 - n} : x = \frac{ln ( - \frac{1}{1 - n} )}{ln ( a )}

x = \frac{ln ( - \frac{1}{1 - n} )}{ln ( a )}

2^{\frac{2}{( l o g _{5} ( x ) )}} = \frac{1}{16}

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

\frac{( 3 2 ^{x - 4} )}{( 4 ^{x - 4} )} = 8^{3 x + 2}

3. 2^{x} + 2^{- 6} = 48

m, 2^{m} 0. 3^{n} 0.72 = 1

What is the answer to (a^x+1)/(a^x)=n ?
The answer to (a^x+1)/(a^x)=n is x=(ln(-1/(1-n)))/(ln(a))