Math Problem Statement

5 log 49

Solution

The expression you provided is:

5log495 \log 49

Using logarithmic properties, you can factor out the constant 5:

5log49=log(495)5 \log 49 = \log(49^5)

Now, calculate 49549^5:

495=(72)5=71049^5 = (7^2)^5 = 7^{10}

So the expression simplifies to:

5log49=log(710)=10log75 \log 49 = \log(7^{10}) = 10 \log 7

Hence, 5log49=10log75 \log 49 = 10 \log 7.

If you need the decimal value of log7\log 7, we can approximate it:

log70.8451\log 7 \approx 0.8451

Thus,

5log4910×0.8451=8.4515 \log 49 \approx 10 \times 0.8451 = 8.451

Would you like further details or have any questions?

Here are 5 related questions you might find useful:

  1. What are the properties of logarithms?
  2. How does the change of base formula work in logarithms?
  3. How do you simplify expressions with logarithmic exponents?
  4. What is the relationship between natural logs and common logs?
  5. How can logarithms be used to solve exponential equations?

Tip: Always remember that logarithms help simplify exponential expressions by turning multiplication into addition and powers into multiplication.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponentiation

Formulas

log(a^b) = b log a

Theorems

Logarithmic Power Rule

Suitable Grade Level

Grades 9-12