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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.PE.79

In Exercises 79–84, solve for y.
79. 3^y = 2^(y+1)

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1
Start with the given equation: \(3^{y} = 2^{y+1}\).
Rewrite the right side using exponent rules: \(2^{y+1} = 2^{y} \cdot 2^{1} = 2^{y} \cdot 2\).
Express the equation as \(3^{y} = 2 \cdot 2^{y}\).
Divide both sides by \$2^{y}$ to isolate terms involving $y$: \(\frac{3^{y}}{2^{y}} = 2\).
Rewrite the left side as a single exponential: \(\left(\frac{3}{2}\right)^{y} = 2\). Then, take the natural logarithm of both sides to solve for \(y\): \(y \cdot \ln\left(\frac{3}{2}\right) = \ln(2)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential Equations

Exponential equations involve variables in the exponent position, such as 3^y or 2^(y+1). Solving these requires techniques to isolate the variable, often by rewriting the equation or applying logarithms.
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Properties of Logarithms

Logarithms are the inverse operations of exponentials and help solve equations where the variable is an exponent. Key properties include log(a^b) = b log(a), which allows bringing down exponents to solve for the variable.
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Equating and Simplifying Exponents

When bases differ and cannot be rewritten to a common base, taking logarithms on both sides allows comparison of exponents. Simplifying the resulting equation leads to isolating the variable and finding its value.
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