Math Problem Statement

9^a = 5, 25^b = 2, 8^c = 3; Find a * b * c.

Solution

The image shows the following system of equations:

  1. 9a=59^a = 5
  2. 25b=225^b = 2
  3. 8c=38^c = 3

You are asked to find the value of abca \cdot b \cdot c.

Step-by-step approach:

To solve for aa, bb, and cc, we can take the logarithm of both sides of each equation.

  1. From 9a=59^a = 5, taking the natural logarithm (or log base 10) on both sides: ln(9a)=ln(5)aln(9)=ln(5)a=ln(5)ln(9)\ln(9^a) = \ln(5) \quad \Rightarrow \quad a \ln(9) = \ln(5) \quad \Rightarrow \quad a = \frac{\ln(5)}{\ln(9)}

  2. From 25b=225^b = 2, taking the logarithm on both sides: ln(25b)=ln(2)bln(25)=ln(2)b=ln(2)ln(25)\ln(25^b) = \ln(2) \quad \Rightarrow \quad b \ln(25) = \ln(2) \quad \Rightarrow \quad b = \frac{\ln(2)}{\ln(25)}

  3. From 8c=38^c = 3, taking the logarithm on both sides: ln(8c)=ln(3)cln(8)=ln(3)c=ln(3)ln(8)\ln(8^c) = \ln(3) \quad \Rightarrow \quad c \ln(8) = \ln(3) \quad \Rightarrow \quad c = \frac{\ln(3)}{\ln(8)}

Now, we calculate the values of aa, bb, and cc using approximate logarithms:

a=ln(5)ln(9)1.60942.19720.732a = \frac{\ln(5)}{\ln(9)} \approx \frac{1.6094}{2.1972} \approx 0.732 b=ln(2)ln(25)0.69313.21890.2154b = \frac{\ln(2)}{\ln(25)} \approx \frac{0.6931}{3.2189} \approx 0.2154 c=ln(3)ln(8)1.09862.07940.5283c = \frac{\ln(3)}{\ln(8)} \approx \frac{1.0986}{2.0794} \approx 0.5283

Finally, we find abca \cdot b \cdot c:

abc0.7320.21540.52830.0833a \cdot b \cdot c \approx 0.732 \cdot 0.2154 \cdot 0.5283 \approx 0.0833

Answer:

abc0.0833a \cdot b \cdot c \approx 0.0833.

Would you like further details or have any questions?

Here are 5 related questions to explore:

  1. How do logarithmic properties simplify the solving of exponential equations?
  2. Why is the natural logarithm often used in solving exponentiation problems?
  3. What is the importance of approximate values in real-life mathematical applications?
  4. How would this approach change if different logarithm bases were used (e.g., base 10)?
  5. Can this method be applied to solve equations involving roots instead of powers?

Tip: When solving exponential equations, taking logarithms is a powerful tool to isolate variables in the exponent.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential Equations
Algebra

Formulas

a = ln(5) / ln(9)
b = ln(2) / ln(25)
c = ln(3) / ln(8)
a * b * c = (ln(5) / ln(9)) * (ln(2) / ln(25)) * (ln(3) / ln(8))

Theorems

Logarithmic Identity: ln(x^n) = n ln(x)

Suitable Grade Level

Grades 9-12