# Help with basic Logarithms

• Nov 15th 2008, 03:16 AM
BG5965
Help with basic Logarithms
Just a side question: Why is Log9(3) = 1/2? Can you please solve it step by step for me. Thanks

1) Solve http://i301.photobucket.com/albums/nn74/BL5965/5a.jpg showing the steps.

2) Solve http://i301.photobucket.com/albums/nn74/BL5965/5c.jpg showing all steps.

Thankyou!
• Nov 15th 2008, 04:11 AM
earboth
Quote:

Originally Posted by BG5965
Just a side question: Why is Log9(3) = 1/2? Can you please solve it step by step for me. Thanks

1) Solve http://i301.photobucket.com/albums/nn74/BL5965/5a.jpg showing the steps.

2) Solve http://i301.photobucket.com/albums/nn74/BL5965/5c.jpg showing all steps.

Thankyou!

To your first question: You are looking for an exponent which belongs to the base 9 such that the result is 3:

$\displaystyle 9^x = 3~\implies~9^x = \sqrt{9} ~\implies~9^x = 9^{\frac12}$

2 powers with the same base are equal if the exponents are equal too. Therefore

$\displaystyle x = \dfrac12$

to #1:

$\displaystyle \dfrac{\log_7(32)}{\log_7(8\sqrt{2}}=\dfrac{5\log_ 7(2)}{3.5\log_7(2)}= \dfrac{10}7$

to #2:

$\displaystyle \dfrac{\log_a(18)+\log_a(12)-\log_a(125)}{\log_a(42) - \log_a(35)} =$$\displaystyle \dfrac{2\log_a(3)+\log_a(2)+2\log_a(2)+\log_a(3)-3\log_a(5)}{\log_a(6)+ \log_a(7) - \log_a(7)-\log_a(5)} =$ $\displaystyle = \dfrac{3\left(\log_a(2)+\log_a(3)-\log_a(5) \right)}{2\log_a(7)+\log_a(6)-\log_a(5) }$
• Nov 15th 2008, 08:34 AM
Soroban
Hello, BG5965!

Quote:

$\displaystyle 2)\;\;\frac{\log_a(18) + \log_a(12) - \log_a(125)}{\log_a(42) - \log_a(35)}$

The numerator is: .$\displaystyle \log_a(2\cdot3^2) + \log_a(2^2\cdot3) - \log_a(5^3)$

. . $\displaystyle = \;\log_a(2) + \log_a(3^2) + \log_a(2^2) + \log_a(3) - \log_a(5^3)$

. . $\displaystyle = \;\log_a(2) + 2\log_a(3) + 2\log_a(2) + \log_a(3) - 3\log_a(5)$

. . $\displaystyle = \;3\log_a(2) + 3\log_a(3) - 3\log_a(5)$

. . $\displaystyle = \;3\bigg[\log_a(2) + \log_a(3) - \log_a(5)\bigg]$

The denominator is: .$\displaystyle \log_a(2\cdot3\cdot7) - \log_a(5\cdot7)$

. . $\displaystyle = \;\bigg[\log_a(2) + \log_a(3) + \log_a(7)\bigg] - \bigg[\log_a(5) + \log_a(7)\bigg]$

. . $\displaystyle = \;\log_a(2) + \log_a(3) - \log_a(5)$

The fraction becomes: .$\displaystyle \frac{3\left[\log_a(2) + \log_a(3) - \log_a(5)\right]} {\log_a(2) + \log_a(3) - \log_a(5)} \;=\;\boxed{3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

An alternate method . . .

We have .$\displaystyle \frac{\log_a(18) + \log_a(12) - \log_a(15)}{\log_a(42) - \log_a(35)} \;= \;\frac{\log_a\left(\dfrac{18\cdot12}{125}\right)} {\log_a\left(\dfrac{42}{35}\right)}$

. . . . . . $\displaystyle = \;\dfrac{\log_a(1.728)}{\log_a(1.2)} \;=\;\log_{1.2}(1.728)$

Since $\displaystyle 1.2^3 \:=\:1.728$, then: .$\displaystyle \log_{1.2}(1.728) \;=\;3$