What is the answer to ln(t)=0.96ln(16)+0.2 ?

The answer to ln(t)=0.96ln(16)+0.2 is t=16^{0.96}e^{0.2}

Solve for t: ln(t)=0.96ln(16)+0.2

The solution is t=16^{0.96}e^{0.2}

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ln (t) = 0.96 ln (16) + 0.2

t = 1 6^{0.96} e^{0.2}

Decimal Notation

t = 17.49097 \dots

Solution steps

ln (t) = 0.96 ln (16) + 0.2

Apply log rules

t = 1 6^{0.96} e^{0.2}

x, ln (x^{2}) = 20

lo g_{9} (x) = \frac{- 3}{2}

lo g_{5 - x} (x^{2} - 2 x + 65) = 2

x, ln (x^{3} y^{3}) = c

x, ln (x^{3} y^{3}) = 1

What is the answer to ln(t)=0.96ln(16)+0.2 ?
The answer to ln(t)=0.96ln(16)+0.2 is t=16^{0.96}e^{0.2}
Solve for t: ln(t)=0.96ln(16)+0.2
The solution is t=16^{0.96}e^{0.2}