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Chapter 1: Problem 8

How is the property \(b^{x+y}=b^{x} b^{y}\) related to the property \(\log _{b}(x y)=\log _{b} x+\log _{b} y ?\)

Short Answer

Expert verified
#Answer# The properties \(b^{x+y} = b^x b^y\) and \(\log_b(xy) = \log_b x + \log_b y\) are related because applying the logarithm to the powers property results in the logarithmic property. We showed this by transforming the given powers property into the logarithmic property through a series of steps: applying logarithms to both sides, using properties of logarithms, simplifying, and rewriting in terms of powers.

Step by step solution

01

Apply logarithm on both sides of the powers property

Let's begin by applying a logarithm with base \(b\) to both sides of the given property of powers: \(b^{x+y}=b^{x}b^{y}\). We will apply \(\log_b\) on both sides: \(\log_b(b^{x+y}) = \log_b(b^x b^y)\)
02

Use properties of logarithms

Now, let's use the properties of logarithms to simplify the expression. Remember that \(\log_b(a^n) = n\log_b(a)\). Therefore we have: \((x+y) \log_b(b) = \log_b(b^x) + \log_b(b^y)\)
03

Simplify and transform the equation

We can simplify the equation further, since we know that \(\log_bb = 1\). So we get: \((x+y) \cdot 1 = x \cdot 1 + y \cdot 1\) \(x + y = x + y\) Now, let's go back to the logarithmic property we wanted to show: \(\log_b(xy) = \log_b x + \log_b y\)
04

Rewrite the logarithmic property to a power property

Using the logarithm definition, we can write the logarithmic property in terms of powers: \(b^{\log_b(xy)} = b^{\log_b x + \log_b y}\)
05

Conclude the relation between the properties

We have shown that when we applied the logarithm to the powers property and simplified it, we arrived at the initial logarithmic property: \(b^{\log_b(xy)} = b^{\log_b x + \log_b y}\) Therefore, the properties \(b^{x+y} = b^x b^y\) and \(\log_b(xy) = \log_b x + \log_b y\) are related. When the logarithm is applied to the powers property, it results in the logarithmic property.

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