Calculate ln(4) with Mixed Base-2 and Base-7 Logarithmic Expression

Question

Calculate the value of the following expression:

ln4×(log7x7log7x4log7x3+log2y4log2y3log2y) \ln4\times(\log_7x^7-\log_7x^4-\log_7x^3+\log_2y^4-\log_2y^3-\log_2y)

Video Solution

Solution Steps

00:00 Solve
00:07 We'll use the logarithm subtraction formula, we'll get the log of their quotient
00:16 Let's use this formula in our exercise
00:52 Let's simplify as much as possible
01:18 Again, we'll use the logarithm subtraction formula
01:33 The logarithm of 1 in any base is always equal to 0
01:43 Let's use this formula in our exercise
01:51 And this is the solution to the question

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Simplify the logarithmic expression using logarithmic identities.
  • Substitute the simplified result back into the main expression and calculate its value.

Now, let's work through each step:

Step 1: Simplify the logarithmic expression. We'll simplify the parts involving log7\log_7 first, then those involving log2\log_2.

For the terms with log7\log_7:
- Convert log7xn\log_7 x^n terms using the power rule: log7x7=7log7x\log_7 x^7 = 7 \log_7 x, log7x4=4log7x\log_7 x^4 = 4 \log_7 x, and log7x3=3log7x\log_7 x^3 = 3 \log_7 x.
- The expression becomes 7log7x4log7x3log7x7 \log_7 x - 4 \log_7 x - 3 \log_7 x.
- Simple arithmetic yields 0log7x0 \log_7 x, which simplifies to 00.

For the terms with log2\log_2:
- Similarly, log2yn\log_2 y^n terms use the power rule: log2y4=4log2y\log_2 y^4 = 4 \log_2 y, log2y3=3log2y\log_2 y^3 = 3 \log_2 y, and log2y=1log2y\log_2 y = 1 \log_2 y.
- The expression is 4log2y3log2y1log2y4 \log_2 y - 3 \log_2 y - 1 \log_2 y.
- Simple arithmetic gives 0log2y0 \log_2 y, which also simplifies to 00.

Step 2: Substitute these back into the original expression:

Original expression:
ln4×(0+0)=ln4×0=0 \ln 4 \times (0 + 0) = \ln 4 \times 0 = 0.

Therefore, the value of the expression is 0 \textbf{0} .

Answer

0 0


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