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Heinemann P2 Questions

Q.1- Solve
log<3>(2x-3)=log<9>(6x^2+19x+2).

Q.2- Prove
If a^x = b^y=(ab)^xy, then
x+y=1.

*note: the numbers written inside<> are bases of logs. It should be smaller than other numbers.

Reply 1

RobbieC

Q.1- Solve
log<3>(2x-3)=log<9>(6x^2+19x+2).

Since on side has a base that is the SQUARE of the other, by squaring the coefficient on the left hand side, we are able to make the right hand side BASE equal to 3.
Therefore:

log<3>(2x-3)^2 = log<3>(6x^2+19x+2)

We can now remove the logs and multiply out the brackets... This gives:

4x^2 - 12x + 9 = 6x^2 + 19x + 2

Gathering like terms, we get:

2x^2 + 31x - 7 = 0

Ummm, you have not written out the question correctly I don't think, since I did this one the other day and it DID factorise... I have a feeling something in there is negative (on the RHS)

But that is the method for solving the question, hope you can do it now!

Reply 2

Fermat

a^x = b^y = (ab)^(xy)

log[(ab)^(xy)]
= ylog[(ab)^x]
= ylog[a^x.b^x]
= ylog[b^y.b^x]
= ylog[b^(x+y)]
= y(x+y)log

log[(ab)^(xy)] = log[b^y] = ylog
y(x+y)log = ylog
=> y(x+y ) = y
(x+y) = 1
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