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Consider the equation 0.5*10^(8t)=73. Solve the equation for t. Express the solution as a logarithm in base-10. t= ◻ Approximate the value of t. Round your answer to the nearest thousandth. t~~

Consider the equation 0.5⋅108t=73 0.5 \cdot 10^{8 t}=73 .\newlineSolve the equation for t t . Express the solution as a logarithm in base−10-10.\newlinet= t= \newline□ \square \newlineApproximate the value of t t . Round your answer to the nearest thousandth.\newlinet≈ t \approx

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Q. Consider the equation 0.5⋅108t=73 0.5 \cdot 10^{8 t}=73 .\newlineSolve the equation for t t . Express the solution as a logarithm in base−10-10.\newlinet= t= \newline□ \square \newlineApproximate the value of t t . Round your answer to the nearest thousandth.\newlinet≈ t \approx
  1. Isolate exponential term: Step 11: Isolate the exponential term.\newlineWe start by isolating the exponential term on one side of the equation. Since the equation is 0.5×108t=730.5\times10^{8t} = 73, we can divide both sides by 0.50.5 to isolate the exponential term.\newline108t=730.510^{8t} = \frac{73}{0.5}\newline108t=14610^{8t} = 146
  2. Apply logarithm: Step 22: Apply logarithm to both sides.\newlineTo solve for tt, we apply the logarithm to both sides of the equation. We use the base−10-10 logarithm.\newlinelog⁡10(108t)=log⁡10(146)\log_{10}(10^{8t}) = \log_{10}(146)\newlineUsing the logarithmic identity log⁡b(bx)=x\log_b(b^x) = x, we simplify the left side.\newline8t=log⁡10(146)8t = \log_{10}(146)
  3. Solve for t: Step 33: Solve for t.\newlineNow, we solve for t by dividing both sides of the equation by 88.\newlinet=log⁡10(146)8t = \frac{\log_{10}(146)}{8}
  4. Approximate value of t: Step 44: Approximate the value of t.\newlineUsing a calculator, we find the value of log⁡10(146)\log_{10}(146) and then divide by 88.\newlinelog⁡10(146)≈2.1644\log_{10}(146) \approx 2.1644\newlinet≈2.1644/8t \approx 2.1644 / 8\newlinet≈0.27055t \approx 0.27055

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