Please help with these log questions?

Question

What is x in the following equations?
1. 1/2log x – log x = log 2
2. [logbase2(x)]/[logbase2(8)]=2/3
3. lnx+ln(x+1)=ln2

I've tried so hard to figure these out. If anyone can help I would appreciate it so much.

Answer 1

first, memorize these rule:

product rule:
log a + log b = log (ab)

quotient rule:
log a - log b = log (a/b)

power rule:
log a^x = x log a

log_a b = x ==> a^x = b

change base rule:
log_b a = loga / logb (b is the base. "_" means base)



1) 1/2logx - logx = log2

power rule:
log √(x) - logx = log2

quotient rule:
log √(x) / x = log2

if log a = log b, then a = b. you have:
√(x) / x = 2

multiply x for both sides
√(x) = 2x

square both sides
x = 4x^2

0 = 4x^2 - x

0 = x (4x - 1)

x = 0 or 1/4

you can't have a 0 for log. THe only solution is 1/4



2) log_2 (x) / log_2 (8) = 2/3

change base rule, you have:
log_2 (x) / 3 = 2/3

multiply 3 for both sides
log_2 (x) = 2

change to exponent form
x = 2^2
x = 4



3) lnx + ln(x + 1) = ln2

product rule
ln (x (x + 1)) = ln2

ln (x^2 + x) = ln2

if lna = lnb then a = b
x^2 + x = 2

x^2 + x -2 = 0

(x + 2) (x - 1) = 0

x = -2 or 1

check
ln (-2 + 1) = ln(-1), you can't have a negarive for log or nateral log. The solution to this problem is 1

hope this helps

Answer 2

1.) Take the whole thing as an exponent of 10 to cancel out the logs.

First, convert 1/2logx
1/2logx = logx^(1/2)

logx^(1/2) – log x = log 2

Then simplify:
logx^(1/2) – log x = log 2
=
(logx^(1/2))/(log x) = log 2

Then take it as an exponent to cancel out the logs.
10^(logx^(1/2))/(log x) = 10^log 2
=
x^(1/2)/x = 2
=
x^(1/2) = 2x
x = 4x^2
4x^2 - x = 0
facotrs to 0 and 1/4
there is no log of 0 so 1/4 is the only answer

I'll skip the second one

3.) lnx+ln(x+1)=ln2
Just make the problem an exponent of e to cancel the logs again.

Simplify:
lnx+ln(x+1)=ln2
=
lnx(x+1) = ln2
Then cancel the logs:
e^lnx(x+1) = e^ln2
=
x(x+1) = 2
solve for x
x^2 + x = 2
x^2 + x - 2 = 0
factor
(x+2)(x-1)
x = 1
x = -2 doesn't work because you can't have the log of a negative.

Hope that helps

*edit*

Sorry, I misread the first problem at first. I fixed my original mistake.

Answer 3

hi
i will help u out in these logarithms question...

i will clear ure basics also....k


Ans1)

Take the whole thing as an exponent of 10 to cancel out the logs.

First, convert 1/2logx
1/2logx = logx^(1/2)

logx^(1/2) – log x = log 2

Then simplify:
logx^(1/2) – log x = log 2
=
(logx^(1/2))/(log x) = log 2

Then take it as an exponent to cancel out the logs.
10^(logx^(1/2))/(log x) = 10^log 2
=
x^(1/2)/x = 2
=
x^(1/2) = 2x
x = 4x^2
4x^2 - x = 0
facotrs to 0 and 1/4
there is no log of 0
so 1/4 is the only answer


Ans-2)
We are given that
[logbase2(x)] / [logbase2(8)]=2

well,
we use the property of logarithms...
now we get,

[log (base 8)2]=2
so,
x=8^2
x=64 is answer

[[[[[[ PROPERTY

=log(base m )n
=log (base z)n / log (base z)m

here,
z can be any positive number (not zero) ]]]]]]]]]]


Ans 3.)

lnx+ln(x+1)=ln2
Just make the problem an exponent of e to cancel the logs again.

Simplify:
lnx+ln(x+1)=ln2
=
lnx(x+1) = ln2
Then cancel the logs:
e^lnx(x+1) = e^ln2
=
x(x+1) = 2
solve for x
x^2 + x = 2
x^2 + x - 2 = 0
factor
(x+2)(x-1)
x = 1
x = -2 doesn't work because you can't have the log of a negative.

so the correct answer is 1.

hope this is the best answer...
all the best..............

Answer 4

1. First, move the 1/2 up into the exponent:
log (x^(1/2)) - log x = log 2
Then combine the left hand terms:
log (x^(1/2)/x) = log 2
log (x^(-1/2)) = log 2
x^(-1/2) = 2
1/sqrt(x) = 2
1 = 2sqrt(x)
sqrt(x) = 1/2
x = 1/sqrt(2) = sqrt(2)/2

2. log(base 2)(8) = 3, so this turns into:
log(base2)(x)/3 = 2/3
log(base2)(x) = 2
2^(log(base2)(x)) = 2^2
x = 2^2 = 4

3. First, combine the left hand logs:
ln(x*(x+1)) = ln 2
e^(ln(x*(x+1))) = e^(ln(2))
x*(x+1) = 2
x^2 + x - 2 = 0
Factoring this, we get:
(x+2)(x-1) = 0
so there are two possible answers: x=-2 and x=1.
However, the -2 answer leads to non-real terms on the left hand side of the original equation, so you may want to discard that one. This leaves you with x=1 as the answer.

Hope that helps!

Answer 5

If x>0
1/2logx-log x= log( x^1/2)/x= log(x^-1/2) =log 2
so x^-1/2=2 and x= 2^-2 =1/4
3) if x>0
ln(x*(x+1)=ln2
so x(x+1) =2 and x^2 +x-2= 0 and x=(-1+3)/2 =1(You must take
the positive solution
2) I can´t see what you wrote in 2)
In any case log base2(8) =3 as 2^3=8