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Given that $a$ and $b$ are positive constants, solve the simultaneous equations a = 3b, \log_3 a + \log_b b = 2 - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2

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Given-that-$a$-and-$b$-are-positive-constants,-solve-the-simultaneous-equations--a-=-3b,-\log_3-a-+-\log_b-b-=-2-Edexcel-A-Level Maths Pure-Question 7-2008-Paper 2.png

Given that $a$ and $b$ are positive constants, solve the simultaneous equations a = 3b, \log_3 a + \log_b b = 2. Give your answers as exact numbers.

Worked Solution & Example Answer:Given that $a$ and $b$ are positive constants, solve the simultaneous equations a = 3b, \log_3 a + \log_b b = 2 - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2

Step 1

a = 3b

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Answer

From the first equation, we have: a=3b.a = 3b.

This shows that aa is three times bb.

Step 2

\log_3 a + \log_b b = 2

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Answer

Now substituting the expression for aa into the second equation:

  1. Rewrite aa:
    log⁡3(3b)+log⁡bb=2.\log_3 (3b) + \log_b b = 2.

  2. Using the property of logarithms, we get:
    log⁡33+log⁡3b+1=2.\log_3 3 + \log_3 b + 1 = 2.

    Since log⁡33=1\log_3 3 = 1, this simplifies to:
    1+log⁡3b+1=2.1 + \log_3 b + 1 = 2.

  3. Rearranging terms results in:
    log⁡3b=0.\log_3 b = 0.

  4. Thus, we find:
    b=30=1.b = 3^0 = 1.

Substituting b=1b = 1 back into the equation for aa:
a=3(1)=3.a = 3(1) = 3.

So, the exact solutions are:
a=3,b=1.a = 3, \quad b = 1.

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