Question

Find the areas between the curves y=2($$log_2(x))/x $$and y=2($$log_4(x))/x $$and the x-axis from x=1 to x=e. What is the ratio of the larger area to the smaller?

Accepted Answer

First, rewrite the logarithmic expressions in terms of natural logarithms to simplify the integrals. Recall that $$ \log_a(x) = \frac{\ln(x)}{\ln(a)} . So$$, express $$ y = \frac{2 \log_2(x)}{x}  as$$ $$ y = \frac{2}{x} \cdot \frac{\ln(x)}{\ln(2)} $$ and $$ y = \frac{2 \log_4(x)}{x}  as$$ $$ y = \frac{2}{x} \cdot \frac{\ln(x)}{\ln(4)} . $$Set up the definite integrals for the areas between each curve and the x-axis from $$ x=1  to$$ $$ x=e . $$The area under each curve is given by $$ A = \int_1^e y \, dx . So$$, write the integrals as $$ A_1 = \int_1^e \frac{2}{x} \cdot \frac{\ln(x)}{\ln(2)} \, dx $$ and $$ A_2 = \int_1^e \frac{2}{x} \cdot \frac{\ln(x)}{\ln(4)} \, dx . $$Factor out constants from the integrals to simplify. For example, $$ A_1 = \frac{2}{\ln(2)} \int_1^e \frac{\ln(x)}{x} \, dx $$ and similarly for $$ A_2 . $$Evaluate the integral $$ \int_1^e \frac{\ln(x)}{x} \, dx . $$Use substitution or recall that $$ \int \frac{\ln(x)}{x} \, dx = \frac{(\ln(x))^2}{2} + C . $$Apply the limits from 1 to $$ e  to $$find the definite integral value. Calculate the ratio of the larger area to the smaller area by dividing the two expressions $$ A_1 $$ and $$ A_2 $$ obtained after integration. Simplify the ratio using properties of logarithms, especially noting that $$ \ln(4) = 2 \ln(2) .$$

Find the areas between the curves y=2(log_2(x))/x and y=2(log_4(x))/x and the x-axis from x=1 to x=e. What is the ratio of the larger area to the smaller?

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Key Concepts

Logarithmic Functions and Change of Base

Definite Integration for Area Calculation

Comparing Areas and Ratios