# Question 4.20: Use the log rules to simplify the following expressions to a......

Use the log rules to simplify the following expressions to a single term, if possible:

(a) $log_b (25) + log_b (70) – log_b (55)$
(b) $4 log_x (7) – 3 log_x (0.85) + log_x (10)$
(c) $12 log_b (12) + 3 log_x (8.25) – 2 log_b (5)$

Then evaluate each expression if b = 3, x = e.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.

(a) $log_b (25) + log_b (70) – log_b (55)$   all logs have the same base

= $log_b (25 × 70) − log_b (55)$       so use rule 1 to add the first two terms

Rule 1     Add      $\log_{b}(M)+\log_{b}(N)\Leftrightarrow\log_{b}(M N)$

= $log_b \left(\frac {25 × 70}{55}\right)$     use rule 2 to divide by the negative term

Rule 2    Subtract     $\log_{b}(M)-\log_{b}(N)\Leftrightarrow\log_{b}\left({\frac{M}{N}}\right)$

= $log_b (31.818)$     this cannot be evaluated unless b is given a value

If b = 3 then using rule 4 to change base 3 to base 10, we have

Rule 4   Change of base      $\log_{b}(N)\Leftrightarrow{\frac{\log_{x}(N)}{\log_{x}(b)}}$

$\log_{3}(31.818)={\frac{\log(31.818)}{\log(3)}}={\frac{1.502\,67}{0.4771}}=3.149\,46$

(b) In $4 log_x (7) – 3 log_x (0.85) + log_x (10)$ start by using rule 3 in reverse, bringing the numbers in that multiply the log term as powers before adding or subtracting using rules 1 and 2. If you look at rules 1 and 2, there are no numbers outside the log terms before adding or subtracting:

$4 log_x (7) – 3 log_x (0.85) + log_x (10)$

$= log_x (7)^4 − log_x (0.85)^3 + log_x (10)$      using rule 3 in reverse

Rule 3   Log of an exponential    $\log_{b}(M^{z})\Leftrightarrow{{{z\log_{b}(M)}}}$

$=\log_{x}\left({\frac{7^{4}}{0.85^{3}}}\right)+\log_{x}(10)$      use rule 2

$= log_x (39 096.275) + log_x (10)$

$= log_x (39 096.275 × 10)$    use rule 1

$= log_x (39 096.275)$

If x = e we obtain In(39 096.275) = 10.573 78.

(c) In this problem there are two different bases, therefore combine terms with the same base only:

$12 log_b(12) + 3 log_x (8.25) – 2 log_b(5)$

$= log_b (12)^{12} + log_x (8.25)^3 − log_b (5)^2$    rule 3 in reverse, bring in the constants as powers

$=log_b \left( \frac {12^{12}}{5^2}\right) + log_x(561.516)$    rule 2 for base b terms

$= log_b(3.566 44 × 10^{11}) + log_x(561.516)$    simplify the numbers

This cannot be simplified to a single term as the bases are different. Given b = 3 and x = e, the expression may be evaluated as

$\log_{3}(3.56644\times10^{11})+\log_{x}(561.516) =\frac{\ln(3.56644\times10^{11})}{\ln(3)}+\ln(561.516)=30.543$

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