Solve the Logarithmic Expression: (log₇6 - log₇1.5)/(3log₇2) × 1/log₍₈₎2

Logarithmic Expressions with Base Changes

log76log71.53log721log82= \frac{\log_76-\log_71.5}{3\log_72}\cdot\frac{1}{\log_{\sqrt{8}}2}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:05 We'll use the logarithm subtraction formula, we'll get log of the numerator
00:19 We'll use this formula in our exercise
00:34 We'll use the power logarithm formula, multiply the number by the coefficient
00:39 We'll use this formula in our exercise
01:04 This is the simplified fraction
01:09 We'll use the formula for 1 divided by log, we'll get the inverse logarithm
01:19 We'll use this formula in our exercise
01:34 We'll use the logarithm division formula
01:39 We'll get the logarithm of the numerator in the base of the denominator
01:49 We'll use this formula in our exercise
01:54 We'll substitute in our exercise and solve
02:09 We'll use the logarithm multiplication formula, we'll switch between the bases
02:29 We'll calculate each logarithm separately, substitute and solve
02:55 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

log76log71.53log721log82= \frac{\log_76-\log_71.5}{3\log_72}\cdot\frac{1}{\log_{\sqrt{8}}2}=

2

Step-by-step solution

To solve this problem, we'll simplify the expression step-by-step, using algebraic rules for logarithms:

  • Step 1: Simplify the numerator log76log71.53log72 \frac{\log_7 6 - \log_7 1.5}{3 \log_7 2}

First, apply the logarithm quotient rule to the numerator:
log76log71.5=log7(61.5)=log74 \log_7 6 - \log_7 1.5 = \log_7 \left(\frac{6}{1.5}\right) = \log_7 4

  • Step 2: Simplify 3log72 3 \log_7 2 in the denominator.

The denominator is 3×log72 3 \times \log_7 2 .

  • Step 3: Address the next part of the expression: 1log82 \frac{1}{\log_{\sqrt{8}} 2} .

By changing the base, use log82=log8212 \log_{\sqrt{8}} 2 = \frac{\log_{8} 2}{\frac{1}{2}} because 8=81/2 \sqrt{8} = 8^{1/2} . Now, log82=13 \log_8 2 = \frac{1}{3} as 81/3=2 8^{1/3} = 2 . So, log82=log281/2=1/31/2=23 \log_{\sqrt{8}} 2 = \frac{\log_2 8}{1/2} = \frac{1/3}{1/2} = \frac{2}{3} .

Therefore, the reciprocal is 1log82=32 \frac{1}{\log_{\sqrt{8}} 2} = \frac{3}{2} .

  • Step 4: Combine and simplify the expression.

The complete logarithmic expression simplifies as follows:
log743log7232=log7(22)3log7232 \frac{\log_7 4}{3 \log_7 2} \cdot \frac{3}{2} = \frac{\log_7 (2^2)}{3 \log_7 2} \cdot \frac{3}{2}

Using the power rule, log74=2log72 \log_7 4 = 2 \log_7 2 . Plug this back into the expression:
2log723log7232 \frac{2 \log_7 2}{3 \log_7 2} \cdot \frac{3}{2}
The log72 \log_7 2 cancels within the fraction, and we are left with 23×32=1 \frac{2}{3} \times \frac{3}{2} = 1 .

Therefore, the solution to the problem is 1 1 .

3

Final Answer

1 1

Key Points to Remember

Essential concepts to master this topic
  • Quotient Rule: logamlogan=loga(mn) \log_a m - \log_a n = \log_a \left(\frac{m}{n}\right) simplifies numerators
  • Power Rule: loga(bc)=clogab \log_a (b^c) = c \log_a b transforms log74=2log72 \log_7 4 = 2 \log_7 2
  • Check: Verify base changes like log82=23 \log_{\sqrt{8}} 2 = \frac{2}{3} using (8)2/3=2 (\sqrt{8})^{2/3} = 2

Common Mistakes

Avoid these frequent errors
  • Incorrectly calculating logarithms with fractional bases
    Don't assume log82=12 \log_{\sqrt{8}} 2 = \frac{1}{2} without proper calculation! This ignores the base change formula and leads to wrong results. Always use loga1/nb=nlogab \log_{a^{1/n}} b = n \log_a b or change to familiar bases first.

Practice Quiz

Test your knowledge with interactive questions

\( \log_{10}3+\log_{10}4= \)

FAQ

Everything you need to know about this question

Why does log76log71.5 \log_7 6 - \log_7 1.5 become log74 \log_7 4 ?

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The quotient rule says logamlogan=loga(mn) \log_a m - \log_a n = \log_a \left(\frac{m}{n}\right) . So log76log71.5=log7(61.5)=log74 \log_7 6 - \log_7 1.5 = \log_7 \left(\frac{6}{1.5}\right) = \log_7 4 .

How do I handle logarithms with bases like 8 \sqrt{8} ?

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Remember that 8=81/2 \sqrt{8} = 8^{1/2} . Use the base change property: loga1/nb=nlogab \log_{a^{1/n}} b = n \log_a b . So log82=2log82 \log_{\sqrt{8}} 2 = 2 \log_8 2 , and since 81/3=2 8^{1/3} = 2 , we get 2×13=23 2 \times \frac{1}{3} = \frac{2}{3} .

Why does log74 \log_7 4 equal 2log72 2 \log_7 2 ?

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This uses the power rule: loga(bc)=clogab \log_a (b^c) = c \log_a b . Since 4=22 4 = 2^2 , we have log74=log7(22)=2log72 \log_7 4 = \log_7 (2^2) = 2 \log_7 2 .

How do I know when logarithms will cancel out?

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Logarithms cancel when you have the same base and argument in numerator and denominator. In our problem, 2log723log72 \frac{2 \log_7 2}{3 \log_7 2} simplifies to 23 \frac{2}{3} because log72 \log_7 2 appears in both.

What's the easiest way to check my final answer?

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For this type of problem, work backwards with your answer. If you got 1, substitute it back and verify each step. Also, try using a calculator to compute the original expression numerically - it should equal your simplified result!

Why is the reciprocal of log82 \log_{\sqrt{8}} 2 equal to 32 \frac{3}{2} ?

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First, log82=23 \log_{\sqrt{8}} 2 = \frac{2}{3} as shown above. The reciprocal is simply 123=32 \frac{1}{\frac{2}{3}} = \frac{3}{2} . This multiplication by 32 \frac{3}{2} helps simplify our final expression.

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