Algebra
Textbooks
Boundless Algebra
Exponents, Logarithms, and Inverse Functions
Working With Logarithms
Algebra Textbooks Boundless Algebra Exponents, Logarithms, and Inverse Functions Working With Logarithms
Algebra Textbooks Boundless Algebra Exponents, Logarithms, and Inverse Functions
Algebra Textbooks Boundless Algebra
Algebra Textbooks
Algebra
Concept Version 11
Created by Boundless

Logarithms of Powers

The logarithm of the pthp\text{th}pth power of a quantity is ppp times the logarithm of the quantity. In symbols, logb(xp)=plogb(x).\log_b(x^p)=p\log_b(x).log​b​​(x​p​​)=plog​b​​(x).

Learning Objective

  • Relate the power rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of powers


Key Points

    • The logarithm of a product is the sum of the logarithms of the factors.
    • An exponent, ppp, signifies that a number is being multiplied by itself ppp number of times. Because the logarithm of a product is the sum of the logarithms of the factors, the logarithm of a number, xxx, to an exponent, ppp, is the same as the logarithm of xxx added together ppp times, so it is equal to plogb(x).p\log_b(x).plog​b​​(x).

Terms

  • base

    A number raised to the power of an exponent.

  • logarithm

    The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.

  • exponent

    The power to which a number, symbol, or expression is to be raised. For example, the 3 in x3x^3x​3​​.


Full Text

The Power Rule for Logarithms

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

 logb(xy)=logb(x)+logb(y)\displaystyle \log _b \left( {xy} \right) = \log _b \left( x \right) + \log _b \left( y \right)log​b​​(xy)=log​b​​(x)+log​b​​(y)

If we apply this rule repeatedly we can devise another rule for simplifying expressions of the form logbxp\log_b x^plog​b​​x​p​​. 

Recall that xpx^px​p​​ can be thought of as x⋅x⋅x⋯xx \cdot x \cdot x \cdots xx⋅x⋅x⋯x where there are ppp factors of xxx. Then we have:

logb(xp)=logb(x⋅x⋯x)=logbx+logbx+⋯+logbx=plogbx\displaystyle \begin{aligned} \log_b(x^p) &= \log_b (x \cdot x \cdots x) \\ &= \log_b x + \log_b x + \cdots +\log_b x \\ &= p\log_b x \end{aligned}​log​b​​(x​p​​)​​​​​=log​b​​(x⋅x⋯x)​=log​b​​x+log​b​​x+⋯+log​b​​x​=plog​b​​x​​ 

Since the ppp factors of xxx are converted to pp p summands by the product rule formula.  

Example 1: Simplify the expression log3(3x⋅9x100)\log_3(3^x\cdot 9x^{100})log​3​​(3​x​​⋅9x​100​​)

First expand the log:

log3(3x⋅9x100)=log3(3x)+log39+log3(x100)\displaystyle \log_3(3^x\cdot 9x^{100}) =\log_3 (3^x) + \log_3 9 + \log_3(x^{100}) log​3​​(3​x​​⋅9x​100​​)=log​3​​(3​x​​)+log​3​​9+log​3​​(x​100​​) 

Next use the product and power rule to simplify:

log3(3x)+log39+log3(x100)=x+2+100log3x\displaystyle \log_3 (3^x) + \log_3 9 + \log_3 (x^{100})= x+2+100\log_3 xlog​3​​(3​x​​)+log​3​​9+log​3​​(x​100​​)=x+2+100log​3​​x

Example 2: Solve 2(x+1)=1032^{(x+1)}=10^32​(x+1)​​=10​3​​ for xxx using logarithms

Start by taking the logarithm with base 222 of both sides:

log2(2(x+1))=log2(103)x+1=3log2(10)x=3log2(10)−1\displaystyle \begin{aligned} \log_2 (2^{(x+1)}) &= \log_2 (10^3)\\ x+1&=3\log_2(10)\\ x&=3\log_2(10)-1 \end{aligned}​log​2​​(2​(x+1)​​)​x+1​x​​​=log​2​​(10​3​​)​=3log​2​​(10)​=3log​2​​(10)−1​​ 

Therefore a solution would be x=3log2(10)−1.x=3\log_2(10) -1. x=3log​2​​(10)−1.

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