# logs

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#1
f(x)=logx (log to base 3)

solve the equation f[3^(x+4)] = 2x basicaly x+4 is the power of 3
0
16 years ago
#2
log 3^(x+4) = 2x

So:

(x+4)log 3 = 2x
4 log 3 = 2x - xlog3

4log 3 = x(2-log3)
x= 4log3/(2-log3)

Get x= 1.253 3dp

Works in original formula so you know it is right. That solution was very toight, don't you think! Dan
0
#3
this doesnt work dan. the answer says log to bas3 (3^x+4) = 2x so x+4=2x therefore x=4
0
16 years ago
#4
If the LHS of the expression is to the base 3 then the problem is fairly straight forward.

In general if Log(base a) c = b

Then a^b = c

As such in this case a=3, b=2x and c=(x+4)
So: 3^2x = 3^(x+4)

As such 2x = x+4
So X=4 QED

Using the first log law. Hope this straightens things out. Dan
0
#5
oooooh ok thanx a lot cheers
0
16 years ago
#6
hiya, could anyone answer this for me:

log(base 3)x - 2log(base x)3 = 1

and

2e^x + 2e^-x = 5
I don't get what you do when there's a minus sign in the power. you can't the substitution y=e^x can you?

I can do all the other log questions, its just those ones!
0
16 years ago
#7
In your first problem change the first log base to x (bases in brackets):

Log(x)x/Log(x)3 - 2Log(x)3 = 1

You now hace question in terms of log(x)

This simplifies to 1/Log(x)s - 2Log(x)3 = 1

Replace Log(x)3 by U

1/U - 2U = 1

Multiply by U
1 - 2U^2 = U

2U^2 + U -1 = 0
(2U-1)(U+1)= 0

So U =1/2 or -1

So Log(x)3= 1/2 (or -1)

x^1/2 = 3 or X^-1 =3

So X = 9 or 1/3

For problem 2)

2e^x + 2e^-x = 5

Multiply through by e^x

2e^2x + 2 = 5e^x

2e^2x - 5e^x + 2 = 0

Replace e^x with y
2y^2 -5y + 2 = 0
(2y-1)(y-2)=0

y=1/2 or 2

e^x = 1/2 or 2
x= ln(1/2) or ln2

Hope these help dude! Dan
0
16 years ago
#8
thanks a million man! had me totally baffled they did! xxx
0
16 years ago
#9
for some reason this was never given in the book, but this is quite an important law in logs. Maybe it can help you.

log(x)y = 1 / log(y)x

I just realised how easy the last question in the P2 exam in January was, when I had no idea how to do it last time.

Simplify 3(2^-log(2)x) - 1

regards,
0
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