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. :smile:

Reply 4

2012_jonathan_s

3

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 :smile:

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.

Reply 6

truyenh

4

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

Reply 7

have

17

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.

Reply 8

monkeyman0121

20

Wow, you bumped a 6-year-old thread! (Since the last comment.) :tongue:

Reply 9

ccmaths

1

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

Reply 10

username3477548

20

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