What is the answer to (1.012^{t-1998})=1.1707 ?

The answer to (1.012^{t-1998})=1.1707 is t=(ln(1.1707))/(ln(1.012))+1998

Solve for t: (1.012^{t-1998})=1.1707

The solution is t=(ln(1.1707))/(ln(1.012))+1998

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(1.01 2^{t - 1998}) = 1.1707

t = \frac{ln ( 1.1707 )}{ln ( 1.012 )} + 1998

Decimal

t = 2011.21213 \dots

Solution steps

(1.01 2^{t - 1998}) = 1.1707

Apply exponent rules

(t - 1998) ln (1.012) = ln (1.1707)

Solve

(t - 1998) ln (1.012) = ln (1.1707) : t = \frac{ln ( 1.1707 )}{ln ( 1.012 )} + 1998

t = \frac{ln ( 1.1707 )}{ln ( 1.012 )} + 1998

2^{4 x - 2} = 4^{x + 2}

5^{3 x + 8} = 2 5^{2 x}

2^{- 3 p} \cdot 2^{2 p} = 2^{2 p}

0.0 1^{x} = 100

\frac{1}{3} e^{3 t} = \frac{1}{2}

What is the answer to (1.012^{t-1998})=1.1707 ?
The answer to (1.012^{t-1998})=1.1707 is t=(ln(1.1707))/(ln(1.012))+1998
Solve for t: (1.012^{t-1998})=1.1707
The solution is t=(ln(1.1707))/(ln(1.012))+1998