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Given $\log_a 2 = p$ and $\log_a 3 = q$, where $a > 0$, write each of the following in terms of $p$ and $q$: (i) $\frac{8}{\log_a 3}$ - Leaving Cert Mathematics - Question b - 2016

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Given-$\log_a-2-=-p$-and-$\log_a-3-=-q$,-where-$a->-0$,-write-each-of-the-following-in-terms-of-$p$-and-$q$:--(i)-$\frac{8}{\log_a-3}$-Leaving Cert Mathematics-Question b-2016.png

Given $\log_a 2 = p$ and $\log_a 3 = q$, where $a > 0$, write each of the following in terms of $p$ and $q$: (i) $\frac{8}{\log_a 3}$. (ii) $\log_a \frac{9a^2}{16}... show full transcript

Worked Solution & Example Answer:Given $\log_a 2 = p$ and $\log_a 3 = q$, where $a > 0$, write each of the following in terms of $p$ and $q$: (i) $\frac{8}{\log_a 3}$ - Leaving Cert Mathematics - Question b - 2016

Step 1

(i) $\frac{8}{\log_a 3}$

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Answer

To express 8loga3\frac{8}{\log_a 3} in terms of pp and qq, we can start with the known logarithm.

Since loga3=q\log_a 3 = q, we can rewrite the expression as follows:

8loga3=8q\frac{8}{\log_a 3} = \frac{8}{q}

Next, we know that loga8\log_a 8 can be expressed in terms of loga2\log_a 2:

loga8=loga(23)=3loga2=3p\log_a 8 = \log_a (2^3) = 3\log_a 2 = 3p

Therefore, we can write:

8loga3=3pq\frac{8}{\log_a 3} = 3p - q

Step 2

(ii) $\log_a \frac{9a^2}{16}$

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Answer

Using the properties of logarithms:

First, we split the logarithm:

loga9a216=loga(9a2)loga(16)\log_a \frac{9a^2}{16} = \log_a(9a^2) - \log_a(16)

Next, we express each term separately:

  1. For loga(9a2)\log_a(9a^2):
    • This can be split as: loga9+logaa2\log_a 9 + \log_a a^2
    • From loga9\log_a 9, since 9=329 = 3^2, we have: loga9=loga(32)=2loga3=2q\log_a 9 = \log_a(3^2) = 2\log_a 3 = 2q
    • And since logaa2=2\log_a a^2 = 2 (that is, the logarithm of a number to its own base): logaa2=2\log_a a^2 = 2

Combining these gives us:

loga(9a2)=2q+2\log_a(9a^2) = 2q + 2

  1. For loga(16)\log_a(16):
    • Since 16=2416 = 2^4, we have: loga16=loga(24)=4loga2=4p\log_a 16 = \log_a(2^4) = 4\log_a 2 = 4p

Combining the two parts:

loga9a216=(2q+2)4p=2q+24p\log_a \frac{9a^2}{16} = (2q + 2) - 4p = 2q + 2 - 4p

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