Math Problem Statement

A+ BLOG-SCHOOL SCIENCE FAIR-SUB DISTRICT MATHS TALENT SEARCH EXAM-HSS.pdf

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Solve from the 4th question

Solution

Let's solve question 4 from Part A of the exam:

Question 4

If 2x=32^x = 3, 3y=53^y = 5, and 10z=810^z = 8, express zz in terms of xx and yy.

Solution:

We start with the given equations:

  1. 2x=32^x = 3
  2. 3y=53^y = 5
  3. 10z=810^z = 8

Taking the logarithm on both sides of each equation:

  1. xlog2=log3x \cdot \log 2 = \log 3 x=log3log2\Rightarrow x = \frac{\log 3}{\log 2}
  2. ylog3=log5y \cdot \log 3 = \log 5 y=log5log3\Rightarrow y = \frac{\log 5}{\log 3}
  3. zlog10=log8z \cdot \log 10 = \log 8 z=log8\Rightarrow z = \log 8 (since log10=1\log 10 = 1).

Now, log8=log(23)=3log2\log 8 = \log(2^3) = 3 \cdot \log 2. Hence, z=3log2z = 3 \cdot \log 2.

Substitute log2\log 2 in terms of xx:

From x=log3log2x = \frac{\log 3}{\log 2}, we get log2=log3x\log 2 = \frac{\log 3}{x}.

Substitute into zz:

z=3log3x.z = 3 \cdot \frac{\log 3}{x}.

Expressing xx in terms of yy:

From y=log5log3y = \frac{\log 5}{\log 3}, we get log3=log5y\log 3 = \frac{\log 5}{y}.

Substitute log3\log 3 into zz:

z=3log5yx.z = 3 \cdot \frac{\frac{\log 5}{y}}{x}.

Thus, the final expression for zz is:

z=3log5xy.z = \frac{3 \cdot \log 5}{x \cdot y}.

This is zz expressed in terms of xx and yy.


Would you like me to simplify this further or solve another question?

Here are related questions for exploration:

  1. Can you express xx and yy independently in terms of zz?
  2. What happens if zz has a different base?
  3. How would zz change if 10z=910^z = 9?
  4. Can you verify these logarithmic transformations numerically?
  5. How would you graph these equations?

Tip: Always simplify logarithmic expressions to make substitution easier.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential Equations
Algebraic Manipulation

Formulas

logarithmic property: log(a^b) = b * log(a)
change of base formula: log_a(b) = log(b) / log(a)

Theorems

Properties of Logarithms
Exponent-Logarithm Relationship

Suitable Grade Level

Grades 11-12