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Precalculus Mathematics for Calculus
Found in: Page 351
Precalculus Mathematics for Calculus

Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

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Short Answer

The function f(x) = log9 x is the logarithm function with base

____. So, width="46" height="21" role="math" style="max-width: none; vertical-align: -5px;" localid="1648528874754" f9= _____, f1= ____, f19= ____, f81= ____, and f3= ____.

The function f(x)=log9x is the logarithm function with base 9. So, f(9)=1,  f(1)=0,  f19=1,  f(81)=2,  and   f(3)=12.

See the step by step solution

Step by Step Solution

Step 1. Given.

The given function is f(x)=log9x.

Step 2. To determine.

We have to find f(9),  f(1),  f19,  f(81),  f(3).

Step 3. Calculation.

We compare f(x)=log9x with f(x)=logax.

Here, a=9.

So, this given function f(x)=log9x has base =9.

We’ll use the formula: log9k=k.

width="92" height="80" role="math" style="max-width: none; vertical-align: -41px;" localid="1648529347686" f(9)=log99 =log991 =1

f(1)=log91 =log990 =0

f19=log919 =log991 =1

f81=log981 =log992 =2

f3=log93 =log9912 =12

Hence, the required values are f(9)=1,  f(1)=0,  f19=1,  f(81)=2,  f(3)=12.

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