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Consider the equation 11*10^(5t)=20". " Solve the equation for t. Express the solution as a logarithm in base10. t= Approximate the value of t. Round your answer to the nearest thousandth. t~~

Consider the equation\newline11⋅105t=20.  11 \cdot 10^{5 t}=20 \text {. } \newlineSolve the equation for t t . Express the solution as a logarithm in base1010.\newlinet= t= \newlineApproximate the value of t t . Round your answer to the nearest thousandth.\newlinet≈ t \approx

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Q. Consider the equation\newline11⋅105t=20.  11 \cdot 10^{5 t}=20 \text {. } \newlineSolve the equation for t t . Express the solution as a logarithm in base1010.\newlinet= t= \newlineApproximate the value of t t . Round your answer to the nearest thousandth.\newlinet≈ t \approx
  1. Isolate exponential term: Isolate the exponential term.\newlineTo solve for tt, we first need to isolate the term with the exponent. We do this by dividing both sides of the equation by 1111.\newline11⋅105t=2011 \cdot 10^{5t} = 20\newline105t=201110^{5t} = \frac{20}{11}
  2. Apply logarithm to both sides: Apply the logarithm to both sides of the equation.\newlineTo solve for the exponent, we apply the logarithm to both sides of the equation. We will use the common logarithm (base 1010).\newlinelog⁡(105t)=log⁡(2011)\log(10^{5t}) = \log\left(\frac{20}{11}\right)
  3. Use property of logarithms: Use the property of logarithms to bring down the exponent.\newlineThe property of logarithms that we use here is log⁡(bx)=x⋅log⁡(b)\log(b^x) = x \cdot \log(b). We apply this property to the left side of the equation.\newline5t⋅log⁡(10)=log⁡(2011)5t \cdot \log(10) = \log\left(\frac{20}{11}\right)
  4. Simplify left side: Simplify the left side of the equation.\newlineSince log⁡(10)\log(10) is equal to 11, the equation simplifies to:\newline5t=log⁡(2011)5t = \log\left(\frac{20}{11}\right)
  5. Solve for t: Solve for t.\newlineTo solve for t, we divide both sides of the equation by 55.\newlinet=log⁡(2011)5t = \frac{\log(\frac{20}{11})}{5}
  6. Approximate value of t: Approximate the value of t.\newlineUsing a calculator, we can find the approximate value of t by evaluating the logarithm and dividing by 55.\newlinet≈log⁡(2011)5 t \approx \frac{\log(\frac{20}{11})}{5} \newlinet≈log⁡(1.81818181818)5 t \approx \frac{\log(1.81818181818)}{5} \newlinet≈0.25963731055 t \approx \frac{0.2596373105}{5} \newlinet≈0.0519274621 t \approx 0.0519274621 \newlineRounded to the nearest thousandth, t≈0.052 t \approx 0.052

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