Math Problem Statement

²log ¹/8 + ³log ¹/9 + ⁵log ¹/125=

Solution

To solve the expression 2log(18)+3log(19)+5log(1125)2\log\left(\frac{1}{8}\right) + 3\log\left(\frac{1}{9}\right) + 5\log\left(\frac{1}{125}\right), let's break it down step by step.

Step 1: Use the logarithm power rule

The power rule of logarithms states that alog(b)=log(ba)a\log(b) = \log(b^a). Applying this rule:

2log(18)=log((18)2)2\log\left(\frac{1}{8}\right) = \log\left(\left(\frac{1}{8}\right)^2\right) 3log(19)=log((19)3)3\log\left(\frac{1}{9}\right) = \log\left(\left(\frac{1}{9}\right)^3\right) 5log(1125)=log((1125)5)5\log\left(\frac{1}{125}\right) = \log\left(\left(\frac{1}{125}\right)^5\right)

Step 2: Simplify the exponents

Now, calculate the values:

(18)2=164\left(\frac{1}{8}\right)^2 = \frac{1}{64} (19)3=1729\left(\frac{1}{9}\right)^3 = \frac{1}{729} (1125)5=131252=19765625\left(\frac{1}{125}\right)^5 = \frac{1}{3125^2} = \frac{1}{9765625}

So, the expression becomes:

log(164)+log(1729)+log(19765625)\log\left(\frac{1}{64}\right) + \log\left(\frac{1}{729}\right) + \log\left(\frac{1}{9765625}\right)

Step 3: Use the logarithm addition rule

The addition rule for logarithms states that log(a)+log(b)=log(a×b)\log(a) + \log(b) = \log(a \times b). Thus, we have:

log(164×1729×19765625)\log\left(\frac{1}{64} \times \frac{1}{729} \times \frac{1}{9765625}\right)

Step 4: Multiply the fractions

Multiplying the fractions:

164×1729×19765625=164×729×9765625\frac{1}{64} \times \frac{1}{729} \times \frac{1}{9765625} = \frac{1}{64 \times 729 \times 9765625}

Calculate the product:

64×729=4665664 \times 729 = 46656 46656×9765625=45580359687546656 \times 9765625 = 455803596875

Thus:

1455803596875\frac{1}{455803596875}

The expression becomes:

log(1455803596875)\log\left(\frac{1}{455803596875}\right)

Step 5: Simplify using properties of logarithms

log(1x)=log(x)\log\left(\frac{1}{x}\right) = -\log(x)

So:

log(1455803596875)=log(455803596875)\log\left(\frac{1}{455803596875}\right) = -\log\left(455803596875\right)

Thus, the final result is:

log(455803596875)-\log\left(455803596875\right)

This is the simplified answer.

Would you like more details or have any questions?


Here are 8 questions to explore this topic further:

  1. How can you use the change of base formula with logarithms?
  2. What is the significance of the logarithm power rule in solving equations?
  3. How can the logarithm product rule be applied to simplify expressions?
  4. What are some practical applications of logarithms in real-world scenarios?
  5. How do different logarithmic bases affect the calculation process?
  6. Can you explain the relationship between logarithms and exponents in detail?
  7. How do logarithms help in solving exponential equations?
  8. What are the common logarithmic identities that are useful in algebra?

Tip: When dealing with logarithms, always remember the key properties such as the product, quotient, and power rules, as they simplify complex expressions significantly.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Logarithmic Rules
Exponents

Formulas

Logarithm power rule
Logarithm addition rule
Logarithm properties

Theorems

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Suitable Grade Level

Grades 10-12