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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 10: Q. 302 (page 1054)

In the following exercises, solve for $x$
$\log_{5} (x + 3) + \log_{5} (x - 6) = \log_{5} 10$

Short Answer

Expert verified

By solving the equation $\log_{5} (x + 3) + \log_{5} (x - 6) = \log_{5} 10$ , the value of $x$ is $7$

Step by step solution

01

Step 1. Given

An expression $\log_{5} (x + 3) + \log_{5} (x - 6) = \log_{5} 10$

To solve the expression for $x$

02

Step 2. Use the Product property

By Product Property of logarithm,

$\log_{a} M + \log_{a} N = \log_{a} M N$

$\log_{5} ((x + 3) (x - 6)) = \log_{5} 10 \log_{5} (x^{2} - 3 x - 18) = \log_{5} 10$

03

Step 3. Use One to One property

If $\log_{a} M = \log_{a} N$ , then $M = N$

$x^{2} - 3 x - 18 = 10 x^{2} - 3 x - 28 = 0$

04

Step 4. Solve the equation

$x^{2} - 3 x - 28 = 0 (x - 7) (x + 4) = 0 x = - 4, 7$

Logarithm for a negative number does not exist.

So, $x = 7$

05

Step 5. Check the solution

Substitute $x = 7$ in the original equation and solve.

$\log_{5} (7 + 3) + \log_{5} (7 - 6) = \log_{5} 10 \log_{5} (10) + \log_{5} (1) = \log_{5} 10 10 = 10$

Hence the solution is checked.

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