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Solving log equations
By Anonymous on Wednesday, April 18, 2001 - 01:07 pm:

I'm having some problems with the following questions. Please help!

  1. Solve the equation lg(5x+8) + lg(2x+1) - 2lg(x+1)
  2. Solve the equation ln(5x+6)= 2ln(5x-6)
  3. Prove that if ax=by=(ab)xy, then x+y=1

Can you please show your working. Thanks in advance
By Sean Hartnoll (Sah40) on Wednesday, April 18, 2001 - 03:25 pm:

There isn't an = sign in the first 'equation'. All these questions involve standard properties of logs. In particular

log(a) + log(b) = log(ab)
log(a) - log(b) = log(a/b)
log(a) = 0 if and only if a = 1
n log(a) = log(an).

Use these to solve the equations by grouping everything together. The generic case is

log(ax + b) = n log(cx+d)

becomes

log((ax+b)/(cx+d)n) = 0

So we need

(ax+b)/(cx+d)n = 1

or

ax+b = (cx+d)n

This should now be some kind of equation that you know how to solve (say if n=2).

Sean

By Anonymous on Wednesday, April 18, 2001 - 05:15 pm:

Thanks Sean.
Could you answer question 3) though. I can't get my head round it.

By Kerwin Hui (Kwkh2) on Wednesday, April 18, 2001 - 05:29 pm:

3) is not true: take x=y=0, a,b to be non-zero.

Kerwin

By Anonymous on Wednesday, April 18, 2001 - 05:37 pm:

If it's not true how can it be an exam question?

By Sean Hartnoll (Sah40) on Wednesday, April 18, 2001 - 06:16 pm:

It is probably assuming that x and y are not zero, in fact this needs to be the case if you want to take logs (the log of zero is not defined). Taking logs:

x log(a) = y log(b) = xy log(ab)
= xy(loga + logb) = y2 log(b) + xy logb
= y(y+x) logb

so then y+x=1 for this to be consistent.

Sean

By Anonymous on Thursday, April 19, 2001 - 01:16 pm:

thanks sean for the help

By Michael Doré (Michael) on Wednesday, April 25, 2001 - 03:07 pm:

It's not the fact that log 0 is undefined that gives us problems. (After all, we aren't taking the log of x,y anywhere.) Rather it is the fact that in Sean's solution we get:

y log(b) = y(y+x) log(b)

Assuming a,b are not 1, we obtain:

y = y(y+x)

So x + y = 1 or y = 0. If y = 0 then x = 0 (again assuming that a,b are not 1), so the two possibilities are x = y = 0 or x + y = 1.