Solve: Combined Logarithmic Fractions with Bases 7, 4, and 2

2log78log74+1log43×log29= \frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29=

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Step-by-step video solution

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00:00 Solve
00:05 We'll use the formula for logarithmic division
00:10 We'll get the log of the numerator in base of the denominator
00:13 To divide 1 by log, we'll get the inverse log in number and base
00:21 We'll calculate the log separately and substitute in the exercise
00:46 We'll use the formula for logarithmic multiplication, switch between bases
01:01 We'll simplify what we can
01:06 We'll calculate each log separately and substitute in the exercise
01:36 And this is the solution to the question

Step-by-step written solution

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1

Understand the problem

2log78log74+1log43×log29= \frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29=

2

Step-by-step solution

To solve the problem 2log78log74+1log43×log29\frac{2\log_7 8}{\log_7 4} + \frac{1}{\log_4 3} \times \log_2 9, we will apply various logarithmic rules:

Step 1: Simplify 2log78log74\frac{2\log_7 8}{\log_7 4}.

  • Using the power property, log78=log723=3log72\log_7 8 = \log_7 2^3 = 3\log_7 2.
  • Similarly, log74=log722=2log72\log_7 4 = \log_7 2^2 = 2\log_7 2.
  • The expression becomes 2×3log722log72=3\frac{2 \times 3\log_7 2}{2\log_7 2} = 3.

Step 2: Simplify 1log43×log29\frac{1}{\log_4 3} \times \log_2 9.

  • 1log43=log34\frac{1}{\log_4 3} = \log_3 4, by inversion.
  • log29\log_2 9 can be expressed as log232=2log23\log_2 3^2 = 2\log_2 3.
  • The product becomes log34×2log23=2log24log23×log23\log_3 4 \times 2\log_2 3 = 2 \cdot \frac{\log_2 4}{\log_2 3} \times \log_2 3.
  • Since log24=2\log_2 4 = 2, this simplifies to 2×21=42 \times \frac{2}{1} = 4.

Step 3: Add the results from Steps 1 and 2:
3+4=73 + 4 = 7.

Therefore, the solution to the problem is 77.

3

Final Answer

7 7

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\( \frac{1}{\log_49}= \)

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