Logarithms and surds

Jun 2010
9
1
Thanks for all your help. I appreciate it a lot.

Question 1
If \(\displaystyle a^2 + b^2 = 7ab\) show that

\(\displaystyle 2 \log_{10} \frac{a+b}{3} = \log_{10} a + \log_{10} b \)

Question 2
If \(\displaystyle 1 + \log_{a} (7x - 3a) = 2 \log_{a} x + \log_{a} 2 \) find in terms of a the possible values of x
 
Jun 2009
806
275
Thanks for all your help. I appreciate it a lot.

Question 1
If \(\displaystyle a^2 + b^2 = 7ab\) show that

\(\displaystyle 2 \log_{10} \frac{a+b}{3} = \log_{10} a + \log_{10} b \)

Question 2
If \(\displaystyle 1 + \log_{a} (7x - 3a) = 2 \log_{a} x + \log_{a} 2 \) find in terms of a the possible values of x
Q.1 \(\displaystyle a^2 + b^2 = 7ab\)

Add 2ab to both the sides. Then

\(\displaystyle a^2 + b^2 + 2ab = 9ab\)

\(\displaystyle (a+b)^2 = 9ab\)

\(\displaystyle (\frac{a+b}{3})^2 = ab\)

Now take log on both the side to the base 10.

Q.2

\(\displaystyle 1 + \log_{a} (7x - 3a) = 2 \log_{a} x + \log_{a} 2 \)


\(\displaystyle \log_{a}(a) + \log_{a} (7x - 3a) = \log_{a}2x^2\)


\(\displaystyle \log_{a}a(7x - 3a) = \log_{a}2x^2\)

\(\displaystyle 2x^2 - 7ax + 3a^2 = 0\)

Now solve the quadratic to find x in terms of a.
 
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