1. ## logs again

I really am trying to understand this. According to your last advice, letting y=ln(x), see if I am getting any closer.

lnx = 2(ln3-ln5)
y = 2(3y-5y)
y = -2y * 2

so (hopefully)
ln(x) = -2e^2

or is it ln(x) = 2e^-2

2. Originally Posted by startingover
I really am trying to understand this. According to your last advice, letting y=ln(x), see if I am getting any closer.

lnx = 2(ln3-ln5)
y = 2(3y-5y)
y = -2y * 2

so (hopefully)
ln(x) = -2e^2

or is it ln(x) = 2e^-2
i'm sure no one here gave you such advice for a problem of this nature. use the laws of logarithms to write the right side as a single log, then just equate what's being logged.

lnx = 2(ln3 - ln5)
=> lnx = 2(ln(3/5)) ...........since logx - logy = log(x/y)
=> lnx = ln(3/5)^2 ...........since log(x^n) = nlogx
=> lnx = ln[(3/5)^2]
=> x = (3/5)^2
=> x = 9/25

besides, even if you wanted to make lnx = y to make things look easier, you would not change ln3 and ln5 into y since they are not lnx, so going your way, you would end up with:

y = 2(ln3 - ln5) ........doing the same operations as above, you'd end up with
y = ln[(3/5)^2]
but y = lnx
=> lnx = ln[(3/5)^2]
=> x = 9/25

3. ## Now I am confused

Quote:
Originally Posted by startingover
5= 3lnx - 1/2lnx

I assume you have to get both sides of the equation with the same term first. If I multiply by lnx, wouldn't it cancel out on the right side? The same for if I divide?

Look at it this way:
Let y = ln(x)
Then:
5 = 3y - (1/2)y

5 = (5/2)y

y = 5*2/5 = 2

ln(x) = 2

x = e^2

-Dan

4. Originally Posted by startingover
Quote:
Originally Posted by startingover
5= 3lnx - 1/2lnx

I assume you have to get both sides of the equation with the same term first. If I multiply by lnx, wouldn't it cancel out on the right side? The same for if I divide?

Look at it this way:
Let y = ln(x)
Then:
5 = 3y - (1/2)y

5 = (5/2)y

y = 5*2/5 = 2

ln(x) = 2

x = e^2

-Dan
ok, so you notice you had 2 lnx's here, in the problem you are doing now, you only have one. if you let y = lnx, you would only change the left side. apparently Dan thought you would have a problem moving lnx around so he made it easier on you by replacing it with a variable, but in general this is not necessary, at least for a problem like this:

5 = 3lnx - 1/2lnx
=> 5 = ln(x^3) - ln(x^(1/2))
=> 5 = ln[(x^3)/(x^(1/2))]
=> 5 = ln[x^(5/2)]
=> e^5 = x^(5/2)
=> (e^5)^(2/5) = x
=> e^2 = x

or you can do it Soroban's way, which i like better, it's not as complicated

you see here that doing it without changing lnx to y makes the problem a bit more challenging, but it still was not necessary. in any case, this is not a problem like the one you have here, on the current problem, there is only one variable ln(x), in the above problem, all variables are lnx, so making a change might make sense if you want a little simplier algebra. so i am still correct about no one telling you that advice for a problem of this nature

usually, we use Dan's technique with slightly harder problems such as quadratics where the variable is a log or another complicated function. example:

(lnx)^2 - 5lnx - 6 = 0 ...............this looks hard, but let y = lnx and we get:
y^2 - 5y - 6 = 0 ......................this looks easy, it's just your old friend the quadratic, but you do the same manipulations