3 logrithmic questions?

Question

Solve for X:

1) log(x) + log(x-3) = 1

2) e^(3k) = 5

3) ln(y) = 2t - 3


I thought I had a handle on this log stuff, but i guess not or are these problems just really complex??

Answer 1

1. first thing you have to apply the addition rule. I am going to use base 10 because we do not have any way to use subscripts. and unless it is specified it is understood to be base 10
addition rule. logX*Y =logX + log Y
using addition rule:
logX + log(X-3) = 1
log X*(X-3) = 1
using equivalency rule
log X= Y = remember using base 10: 10^Y = X

that is going to give me:

10^1=X*(X - 3)
10 = x^2 - 3X
which gives me in turn by transposing the ten.

X^2 -3X - 10 = 0
which is a quadratic equation so to find X you have to factor,

that comes out to

(X +2)(X-3) =0 after factoring
so X is going to equal to:
X=2
X= -3

2. if K stands for Kilo then 3K means 3*1,000 or 3,000 simple mutlicative rule there.

which gives you.

e^(3K) = 5
and is written in exponential form.
to write in log form simply use equivalency rule

which is
In(5) = 3k
take antilog of 5
1.61 = 3K
transpose and divide
K = 1.61/3
K = .536
round off
K = 0.5 or one half.
so K did not stand for Kilo or 1,000

3. In(y) = 2t -3 is already in log form to rewrite it to exp. form use your equivalance rule.

LogX = Y = using base 10: ! 10^Y = X

now you have:

In(y) = 2t-3
rewriting in exponential form gives you:

e^(2t - 3) = y
Now if your professor wants you to carry it further go back and review making bases match in exponents. Don't know if you have gotten into Calculus yet? If you are into Calculus already you can differeniate t with respect to y and y with respecit to t.
Hope that helps.

Answer 2

I assume, you mean log_10(x). You will need the property log(a) + log(b) = log(ab). Also, the domain of any log function is (0, ∞)

1. log(x) + log(x-3) = log(x^2-3x) = 1 hence x^2 -3x = 10^1 = 10. Solving, (x-5)(x+2) = 0 Hence x =5 and x = -2 are possible answers. Note that we must reject x = -2 because log(-2) is undefined. Thus the only answer is x =5

2. e^(3k) = 5. Taking ln of both sides, we get 3k = ln(5). Hence k = ln(5)/3

3. I'm not quite sure what you want to do here, since there are two variables t and y.

Answer 3

For the first one you need to use a property of logs: When you add logs of the same base, you multiply the number out front.

log(x) + log(x-3) = 1
log(x^2-3x) = 1

Now all you do is solve for x:

log(x^2-3x) = 1
x^2 - 3x = 10 >>> Definition of logs
x^2 - 3x -10 = 0
(x+2)(x-5) = 0
x = 5 >>> Since logs can't have negative numbers in front of them.


For the second use the definition of logs:

e^(3k) = 5
3k = e^5
k = (e^5)/3


For the third, it is the definition of logs again:

ln(y) = 2t - 3
y = e^(2t-3)

Answer 4

1) log(x) + log(x-3) = 1

log[x*(x-3)] = 1

x*(x-3) = 10

etc.
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2) e^(3k) = 5

e^(3k) = 5

3k = ln 5

etc.
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3) ln(y) = 2t - 3

y = e^(2t - 3)
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