What is the answer to 4000/2000 =1.25^n ?

The answer to 4000/2000 =1.25^n is n=(ln(2))/(ln(1.25))

Solve for n: 4000/2000 =1.25^n

The solution is n=(ln(2))/(ln(1.25))

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\frac{4000}{2000} = 1.2 5^{n}

n = \frac{ln ( 2 )}{ln ( 1.25 )}

Decimal Notation

n = 3.10628 \dots

Solution steps

\frac{4000}{2000} = 1.2 5^{n}

Apply exponent rules

ln (\frac{4000}{2000}) = n ln (1.25)

Solve

ln (\frac{4000}{2000}) = n ln (1.25) : n = \frac{ln ( 2 )}{ln ( 1.25 )}

n = \frac{ln ( 2 )}{ln ( 1.25 )}

(\frac{7}{5})^{x - 1} = (\frac{5}{7})^{x - 3}

y, x^{4} - 2 x^{y} - 1 = 0

10 t e^{- t} = 0

(\frac{21}{20})^{n} = \frac{27783}{24000}

3^{- 2} \cdot 0. 3^{4} \cdot 0. 3^{n} = 3

What is the answer to 4000/2000 =1.25^n ?
The answer to 4000/2000 =1.25^n is n=(ln(2))/(ln(1.25))
Solve for n: 4000/2000 =1.25^n
The solution is n=(ln(2))/(ln(1.25))