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Chapter 3: Problem 79

Suppose \(b\) and \(y\) are positive numbers, with \(b \neq 1\) and \(b \neq \frac{1}{2} .\) Show that $$ \log _{2 b} y=\frac{\log _{b} y}{1+\log _{b} 2} $$

Short Answer

Expert verified
After applying the change of base formula and properties of logarithms, we can rewrite the given equation as: \[ \frac{\log y}{\log(2b)} = \frac{\log _{b}y}{1+\log_{b}2} \] Further manipulation yields: \[ \frac{\log y}{\log 2 + \log b} = \frac{\log y}{\log 2 + \log b} \] Thus, \(\log_{2b}y = \frac{\log_{b}y}{1+\log_{b}2}\).

Step by step solution

01

Apply Change of Base Formula

Using the change of base formula, which states that \(log_{a}x = \frac{log_{c}x}{log_{c}a}\), where \(c\) is the new base, we can rewrite the given equation as: \[ \frac{\log y}{\log(2b)} = \frac{\log _{b}y}{1+\log_{b}2} \]
02

Rewrite the Denominator on the Left Side

Now, we can rewrite the denominator on the left side using the properties of logarithms. Using the properties, \(\log(ab) = \log a + log b\), we can rewrite the denominator as: \[ \frac{\log y}{\log 2 + \log b} = \frac{\log _{b}y}{1+\log_{b}2} \]
03

Apply Change of Base Formula for the Right Side

We can apply the change of base formula again for the right side of the equation to have the same base as the left side. \[ \frac{\log y}{\log 2 + \log b} = \frac{\frac{\log y}{\log b}}{\frac{\log(2b)}{\log b}} \]
04

Rewrite the Denominator of the Right Side

As we did in step 2, we can rewrite the denominator on the right side using the properties of logarithms. \[ \frac{\log y}{\log 2 + \log b} = \frac{\frac{\log y}{\log b}}{\frac{\log 2 + \log b}{\log b}} \]
05

Simplify the Right Side

Now, we can simplify the right side of the equation. \[ \frac{\log y}{\log 2 + \log b} = \frac{\log y}{\log 2 + \log b} \] This confirms that \(\log_{2b}y = \frac{\log_{b}y}{1+\log_{b}2}\).

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