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$25-34=$ Evaluating Logarithms Evaluate the expression.$$\begin{array}{lll}{\text { (a) } \log _{2} 2} & {\text { (b) } \log _{5} 1} & {\text { (c) } \log _{6} 6^{5}}\end{array}$$

$=5$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 3

Logarithmic Functions

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Lectures

02:22

$25-34=$ Evaluating Logari…

02:21

01:23

Evaluating Logarithms Eval…

02:09

03:29

03:18

00:45

In $27-56,$ evaluate each …

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Evaluate each expression w…

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01:49

01:45

Evaluate each logarithm. <…

00:57

Evaluate each logarithm.

00:40

Evaluate each logarithm. S…

00:46

In Exercises 105–108, eval…

01:18

01:11

01:24

Assume that $\log 4 \appro…

Evaluate expression.$\…

00:27

in this question, we're going to make use off to logarithmic rules. The 1st 1 state that if the base and the value or the same, then the whole log becomes one, right? The other rule that we need here is that lock off a base A with the value B to the power off. See will become. See Look a be right. So take note If this when these two are the same, then the whole lock becomes one. When I have the value to an exponents, the exponents goes, multiply in front and you still have the lock off a b. So in a we have look 22 which from the first rule that we have here because thes two are the same, that will be equal to one then for the lock with based five off one, I can really like the one as five to the power of zero. Then the zero can multiply in front according to the second rule that we use here and that will be zero times one thing because these two are the same, which is zero in lock to the base. Six off, six to the power off five from the second rule, the five can multiply in front. We have log off by 66 which is then equal to one and five times one is five.

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