What is the answer to sqrt(3)^{(n-1)}=1200 ?

The answer to sqrt(3)^{(n-1)}=1200 is n=(2ln(1200))/(ln(3))+1

Solve for n: sqrt(3)^{(n-1)}=1200

The solution is n=(2ln(1200))/(ln(3))+1

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3^{(n - 1)} = 1200

n = \frac{2 ln ( 1200 )}{ln ( 3 )} + 1

Decimal Notation

n = 13.90733 \dots

Solution steps

(3)^{(n - 1)} = 1200

Apply exponent rules

\frac{1}{2} (n - 1) ln (3) = ln (1200)

Solve

\frac{1}{2} (n - 1) ln (3) = ln (1200) : n = \frac{2 ln ( 1200 )}{ln ( 3 )} + 1

n = \frac{2 ln ( 1200 )}{ln ( 3 )} + 1

2^{x + 1} + 2^{x} - 2^{x - 1} = 3^{3} - 7

2^{x + 1} = 3 (5^{x})

\frac{1}{e ^{t}} = e^{- t}

x, x^{2} = 2^{x}

0.08 = \frac{20}{2 ^{x}}

What is the answer to sqrt(3)^{(n-1)}=1200 ?
The answer to sqrt(3)^{(n-1)}=1200 is n=(2ln(1200))/(ln(3))+1
Solve for n: sqrt(3)^{(n-1)}=1200
The solution is n=(2ln(1200))/(ln(3))+1