My Math Forum Lots of Logarithms question

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 November 21st, 2014, 09:35 PM #1 Newbie   Joined: Nov 2014 From: HK Posts: 3 Thanks: 1 Lots of Logarithms question Hi, Even though I know the basic rules for Logarithms, I still can't do the question listed below... can anyone explain how can i solve these problems please? Thanks~ (Assume all base of Logs to be 10) Q1. If log2 = a and log9 = b, then log12 = ? Q2. Let x>y>0. If log(x+y) = a and log(x-y) = b, then log(x^2-y^2)^(1/2) = ? Q3. log2 = a and log3 =b. Express log15 in terms of a and b. Q4. If logx^2 = log3x+1, then x = ? Q5. If 5 = 10^a and 7 = 10^b, then log(7/50) = ? -------------------------------------------------------------------------- Basic rules for Logarithms:
 November 21st, 2014, 10:14 PM #2 Newbie     Joined: Oct 2014 From: Australia Posts: 27 Thanks: 10 Q1: If log a + log b = log (ab) then to find log 12 you'd want to find the factors of 12 as terms of a and b. In this case, 4 and 3 multiply to make 12. If n log a = log (a^n) then we can get log 4 out of log 2 by multiplying it by 2 Similarly we can get log 3 out of log 9 by dividing it by 2 (2 log 3 = log 9) That means that (2 log 2) + ((log 9)/2) should equal log 12 so log 12 = 2a + b/2 Q2: You can factorise (x^2 - y^2) into (x+y)(x-y). log (x^2 - y^2)^(1/2) also equals (1/2)*log (x^2 - y^2) and if you replace it with the factorised version, you get (log (x+y)(x-y))/2 As we already know you log ab = log a + log b, so you get $\displaystyle \frac{\log {(x+y)} + \log {(x-y)}}{2}$ Substitute for the letters and it equals (a+b)/2 Q3: Same as Q1, I'll let you try that one Q4: I'm not sure if this is log (3x+1) or log(3x) + 1, but I'll show both ways. If it's the first one, there's a log on both sides of the equation, so they cancel out. x^2 = 3x+1 and you can solve it using the quadratic formula. If it's the second version, you have to find what 1 is in terms of logs so you can cancel them. log 10 = 1 here so you have log (3x) + log 10. Use the addition rule and you get log x^2 = log 30x. From there you cancel out the logs and solve for x. Q5. If 5 = 10^a then log 5 = a. In the same way, log 7 = b. log (7/50) = log 7 - log 50 and log 50 = log 5 + log 10 If you substitute that it equals log 7 - (log 5 + log 10) log 10 equals 1 therefore log 7 - log 5 - 1 Substitute a and b and it is b-a-1 Thanks from heaven9959
 November 21st, 2014, 10:17 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Q1. 2a + b/2 = log(4) + log(3) = log(12) Q2. (a + b)/2 = 1/2 * log(x² - y²) = log((x² - y²)^(1/2)) Q3. b + 1 - a = log(3) + log(10) - log(2) Q4. $\frac{3+\sqrt{13}}{2}$. Set x² = 3x + 1 and apply the quadratic formula after rearranging. Q5. b - (a + 1). log(7/50) = log(10^b) - log(10^(a + 1)) Thanks from heaven9959 Last edited by greg1313; November 21st, 2014 at 10:32 PM.
November 21st, 2014, 10:17 PM   #4
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Hello, heaven9959!

I'll do a few of them for you.

Quote:
 (Assume all base of Logs to be 10) $\log2 = a$ and $\log9 = b$, find $\log12.$

$\log(12) \;=\;\log(4\cdot3) \;=\;\log(4) + \log(3) \;=\;\log\left(2^2\right) + \log\left(9^{\frac{1}{2}}\right)$

$\qquad =\;2\log(2) + \frac{1}{2}\log(9) \;=\; 2a + \frac{1}{2}b$

Quote:
 Q3. $\log2 = a,\;\log3 =b.\;\;\text{Find }\log15.$

$\log(15) \;=\;\log(3\cdot5) \;=\; \log\left(3\cdot\frac{10}{2}\right)$

$\qquad =\;\log(3) + \log(10) - \log(2) \;=\;b + 1 - a$

Quote:
 Q5. If $5 = 10^a$ and $7 = 10^b$, find $\log\left(\frac{7}{50}\right)$

$5 = 10^a \quad\Rightarrow\quad a = \log(5)$
$7 = 10^b \quad\Rightarrow\quad b = \log(7)$

$\log\left(\frac{7}{50}\right) \;=\; \log(7) - \log(50) \;=\;\log(7) - \log(5\cdot10)$

$\qquad =\;\log(7) - \big[\log(5) + \log(10)\big] \;=\; b - (a + 1)$

November 21st, 2014, 10:26 PM   #5
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Quote:
 Originally Posted by heaven9959 Hi, If log2 = a and log9 = b, then log12 = ?

$\displaystyle \log 12 = \log 4\cdot3=\log4+\log3=\log2^2+\log9^{\frac{1}{2}}=2\ log 2+\dfrac{1}{2}\log 9 = 2a+\dfrac{1}{2}b$

 November 21st, 2014, 10:29 PM #6 Newbie   Joined: Nov 2014 From: HK Posts: 3 Thanks: 1 Thanks guys, I appreciate that so much~ !! Anyway, I still don't get what Question number 4 is doing. Thanks from greg1313 Last edited by heaven9959; November 21st, 2014 at 11:23 PM.
November 22nd, 2014, 05:07 AM   #7
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Quote:
 Originally Posted by heaven9959 Q4. If logx^2 = log3x+1, then x = ?

For the existence of the logarithm, we must have $\displaystyle x > 0$.

$\displaystyle \color{blue}{\log x^2 = \log3x +1\ \Longleftrightarrow\ \log x^2 - \log 3x = 1 \ \Longleftrightarrow\ \log \dfrac{x^2}{3x}=1 \ \Longleftrightarrow\ \log \dfrac{x}{3}=1 \ \Longleftrightarrow\ \\\;\\ \ \Longleftrightarrow\ \log \dfrac{x}{3} = \log 10 \ \Longleftrightarrow\ \dfrac{x}{3} = 10 \ \Longleftrightarrow\ x = 30 . }$

November 23rd, 2014, 02:38 AM   #8
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Math Focus: Trigonometry
Quote:
 Originally Posted by heaven9959 Thanks guys, I appreciate that so much~ !! Anyway, I still don't get what Question number 4 is doing.
x^2 = 3x+1

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# log12=log9 log4

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