Hard Logarithm Question

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Palabras
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#1
Report Thread starter 12 years ago
#1


(\log_3 p)^2 = \log_3(p^2)


\log_3(p+q)=\log_3 p + \log_3 q

Solve these simultaneous equations.

I've got  (\log_3 p)^2 = \log_3 (\frac{p^2 +qp}{q}) so far, but I don't know how to eliminate q.
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DFranklin
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#2
Report 12 years ago
#2
Solving the first equation gives you p, and then it's easy to find q from the 2nd.
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Palabras
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#3
Report Thread starter 12 years ago
#3
Ok thanks, got it.
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id1
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#4
Report 11 years ago
#4
(Original post by Dan12)
How do you find p?
In the first equation, for the RHS, use the logarithmic identity that states that "the logarithm of the n-th power of a number is the same as n times the logarithm of the number." Then, treat the entire equation as a quadratic in log_3p (hint: substitute m=log_3p). Let me know if you still don't get it.
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2012_jonathan_s
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#5
Report 9 years ago
#5
Q1. (log3P)2 = log3(P2)
(log3P)2 = 2(log3P)

Let log3P = x

x2 = 2x
x = 2

log3P = 2
P = 9

Q2. log3(P+Q) = log3P + log3Q
log3(P+Q) = log3(PQ)
P+Q = PQ

Substitute P = 9 from Q1

9+Q = 9Q
9 = 8Q
9/8 = Q
Q= 1 1/8 or 1.125


Therefore, P is 9 and Q is 1.25
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Mr M
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#6
Report 9 years ago
#6
(Original post by 2012_jonathan_s)
...
Welcome to TSR Jonathan.

Two rules you should know.

Don't bump old threads for no reason.

Don't provide full solutions.
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truyenh
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#7
Report 4 years ago
#7
For the first equation x^2 = 2x, there is two solutions for x, x = 0 and x = 2 (where x = log base3 of P)- For x=0 then P=1- For x=2 then P=9However for the second equation P Q=PQ => Q = P/(P-1) - For P=1, Q is undefined so P=1 is rejected as the solution for the whole system. - For P=9, Q = 9/(9-1) = 9/8 = 1.125
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have
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#8
Report 4 years ago
#8
(Original post by truyenh)
For the first equation x^2 = 2x, there is two solutions for x, x = 0 and x = 2 (where x = log base3 of P)- For x=0 then P=1- For x=2 then P=9However for the second equation P Q=PQ => Q = P/(P-1) - For P=1, Q is undefined so P=1 is rejected as the solution for the whole system. - For P=9, Q = 9/(9-1) = 9/8 = 1.125
your madd still. Read the post above you fam.
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monkeyman0121
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#9
Report 4 years ago
#9
Wow, you bumped a 6-year-old thread! (Since the last comment.)
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ccmaths
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#10
Report 1 year ago
#10
forget about q for now- the first equation only has 1 variable and so we can solve without help from the other- so (log3(P))²=log3(P²) which means (log3(P))²=2log2(P) {using log laws, log(x^n) = nlog(x)}. now to get rid of the ² we can divide by log3(P) on both sides, to give us log3(P)=2, applying base 3 to each side, P=3² -> P=9.now we can move onto the second. we’re told log3(P Q)=log3(P) log3(Q), we also know P=9, so log3(P)=log3(9)=2. so we can rewrite this as log3(9 Q)=2 log3(Q). raising each side to base 3, we get 9 Q=3^(2 log3(Q))=3²3^(log3(Q)), the 3^log3 cancels out so we have the equation 9 Q=9Q. now it’s just a simple linear equation to solve, 8Q=9, Q=9/8, so we’re left with the results P=9, Q=9/8
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username3477548
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#11
Report 1 year ago
#11
(Original post by ccmaths)
forget about q for now- the first equation only has 1 variable and so we can solve without help from the other- so (log3(P))²=log3(P²) which means (log3(P))²=2log2(P) {using log laws, log(x^n) = nlog(x)}. now to get rid of the ² we can divide by log3(P) on both sides, to give us log3(P)=2, applying base 3 to each side, P=3² -> P=9.now we can move onto the second. we’re told log3(P Q)=log3(P) log3(Q), we also know P=9, so log3(P)=log3(9)=2. so we can rewrite this as log3(9 Q)=2 log3(Q). raising each side to base 3, we get 9 Q=3^(2 log3(Q))=3²3^(log3(Q)), the 3^log3 cancels out so we have the equation 9 Q=9Q. now it’s just a simple linear equation to solve, 8Q=9, Q=9/8, so we’re left with the results P=9, Q=9/8
thread is 11 years old. OP has probably retired.
Last edited by username3477548; 1 year ago
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haleemazarina
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#12
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#12
dont be rude to jonathon
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haleemazarina
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#13
Report 9 months ago
#13
(Original post by have)
your madd still. Read the post above you fam.
why u typing like that u cretin
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