2log(x-1) + logx = logx + log4

  • Thread starter nae99
  • Start date
I will definitely be more careful in the future!In summary, the conversation is about solving the equation 2log(x-1) + logx = logx + log4 algebraically. The expert summarizer advises to cancel out the same terms on both sides, leaving a log of something equal to a log of something else. The next step is to use the identities p(log_a b) = log_a b^p and log_a u - log_a v = log_a(u/v) to simplify the equation. The resulting equation can then be solved further by using the quadratic formula. However, the expert summarizer suggests taking the square root of both
  • #1
nae99
129
0

Homework Statement



2log(x-1) + logx = logx + log4


Homework Equations





The Attempt at a Solution


log(x-1)^2 + logx = logx (4)
 
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  • #2


Hey there,

ok it looks like you have some exact same term on both sides to begin with, so those can just cancel out right? This leaves you with a log of something = a log of something else, that should simplify it down nice for ya.
 
  • #3


i don't get it
 
  • #4


Omega_Prime said:
Hey there,

ok it looks like you have some exact same term on both sides to begin with, so those can just cancel out right? This leaves you with a log of something = a log of something else, that should simplify it down nice for ya.

so are u saying it should be:
log (2x - x)^2 = 4x
 
  • #5


Well the first step is just an algebra step, you have log (x) on both sides of the = sign, so the equation can be reduced. When that happens you are left with a log (of something) = log (of something else) when this happens you can sort of drop the log and just solve for x.
 
  • #6


nae99 said:
so are u saying it should be:
log (2x - x)^2 = 4x

Noooooo, :)
What you need to do here is just first cancel out the terms [itex]\log x[/itex] on both sides.
Then shift [itex]\log 4[/itex] to the left side remaining 0 at the right side.
Now using these two identities, try to figure out the answer:-
[tex]p(\log_a b)=log_a b^p[/tex]
and
[tex]\log_a u-\log_a v=\log_a \frac{u}{v}[/tex]
 
  • #7


Omega_Prime said:
Well the first step is just an algebra step, you have log (x) on both sides of the = sign, so the equation can be reduced. When that happens you are left with a log (of something) = log (of something else) when this happens you can sort of drop the log and just solve for x.

log (x-1) = log4

i don't think i understand that
 
  • #8


Pranav-Arora said:
Noooooo, :)
What you need to do here is just first cancel out the terms [itex]\log x[/itex] on both sides.
Then shift [itex]\log 4[/itex] to the left side remaining 0 at the right side.
Now using these two identities, try to figure out the answer:-
[tex]p(\log_a b)=log_a b^p[/tex]
and
[tex]\log_a u-\log_a v=\log_a \frac{u}{v}[/tex]

2log (x-1) - log 4 = 0
 
  • #9


nae99 said:
2log (x-1) - log 4 = 0

There you go! I'd have kept log(4) on the right side and then just solve for x.
 
  • #10


nae99 said:
2log (x-1) - log 4 = 0

Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave. :smile:
 
  • #11


Pranav-Arora said:
Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave. :smile:

log (x-1)^2 - log 4 = 0
 
  • #12


nae99 said:
log (x-1)^2 - log 4 = 0

Now use the second identity. :wink:
 
  • #13


Pranav-Arora said:
Now use the second identity. :wink:

log (x^2-1) /4 = 0
 
  • #14


nae99 said:
log (x^2-1) /4 = 0

It would be
[tex]\log \frac{(x-1)^2}{4}=0[/tex]

Whatever the base would be, when you use the antilog method here, the right hand side would become 1. So what you get is:-
[tex]\frac {(x-1)^2}{4}=1[/tex]

Now solve it further. :wink:
 
  • #15


Pranav-Arora said:
It would be
[tex]\log \frac{(x-1)^2}{4}=0[/tex]

Whatever the base would be, when you use the antilog method here, the right hand side would become 1. So what you get is:-
[tex]\frac {(x-1)^2}{4}=1[/tex]

Now solve it further. :wink:

ok so i would do this next but i don't think its right

(x^2-1) /4 = 1
 
  • #16


nae99 said:
ok so i would do this next but i don't think its right

(x^2-1) /4 = 1

I don't understand how you convert (x-1)2 to (x2-1)? :confused:
That's wrong!

Solving this equation:-
[tex]\frac {(x-1)^2}{4}=1[/tex]
we get,
[tex](x-1)^2=4[/tex]
Now try to do it.
 
  • #17


ok, but i thought (x-1) meant (x-1) (x-1)
 
  • #18


nae99 said:
ok, but i thought (x-1) meant (x-1) (x-1)

I still didn't get you :confused:
[tex](x-1)^2=(x-1)(x-1)=x^2-2x+1[/tex]
 
  • #19


Pranav-Arora said:
I still didn't get you :confused:
[tex](x-1)^2=(x-1)(x-1)=x^2-2x+1[/tex]

so i would end up with:

x^2-2x+1 = 4
 
  • #20


and then i would have to do the quadratic equation, right
 
  • #21


nae99 said:
so i would end up with:

x^2-2x+1 = 4

Yes, now proceeding further you get
[tex]x^2-2x-3=0[/tex]
Now this is quadratic equation, try to solve it. :)
 
  • #22


Pranav-Arora said:
Yes, now proceeding further you get
[tex]x^2-2x-3=0[/tex]
Now this is quadratic equation, try to solve it. :)

x = -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] (-2)^2 - 4*1*(-3)[itex]/[/itex] 2*1

x= -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] -4+12[itex]/[/itex] 2

how is that
 
  • #23


nae99 said:
x = -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] (-2)^2 - 4*1*(-3)[itex]/[/itex] 2*1

x= -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] -4+12[itex]/[/itex] 2

how is that

What is the square of (-2)? :smile:
 
  • #24


:confused:i don't know
 
  • #25


Pranav-Arora said:
What is the square of (-2)? :smile:

is it -4
 
  • #26


nae99 said:
is it -4

Nooo, it's 4.
(-2)2=(-2)(-2)=4
 
  • #27


oh ok
 
  • #28


nae99 said:
oh ok

Now what's the answer?
 
  • #29


Pranav-Arora said:
Now what's the answer?

x= -(-2) ± √ 4+12/ 2

x = -(2) [itex]\pm[/itex][itex]\sqrt{}[/itex] 16 [itex]/[/itex] 2

x = -(-2) [itex]\pm[/itex] 4 [itex]/[/itex] 2

x = -(-2) + 4 [itex]/[/itex] 2

x = 6[itex]/[/itex] 2

x= 4
OR

x = -(-2) - 2 [itex]/[/itex] 2

x = -2[itex]/[/itex] 2

x = -1
 
  • #30


nae99 said:
x= -(-2) ± √ 4+12/ 2

x = -(2) [itex]\pm[/itex][itex]\sqrt{}[/itex] 16 [itex]/[/itex] 2

x = -(-2) [itex]\pm[/itex] 4 [itex]/[/itex] 2

x = -(-2) + 4 [itex]/[/itex] 2

x = 6[itex]/[/itex] 2

x= 4
OR

x = -(-2) - 2 [itex]/[/itex] 2

x = -2[itex]/[/itex] 2

x = -1

What are you doing?
6/2=3 and 2-2=0.
Your both the answers are wrong. Correct them.
 
  • #31


Pranav-Arora said:
What are you doing?
6/2=3 and 2-2=0.
Your both the answers are wrong. Correct them.

OMG, such simple mistakes huh, ok

x= 6/2 = 3

x= -(-2) - 2/ 2 = 0/2 x= 0
 
  • #32


That's too much unnecessary work. From,

[tex](x-1)^2=4[/tex]

Take the square root of both sides (remember the [itex]\pm[/itex]).
 
  • #33


Mentallic said:
That's too much unnecessary work. From,

[tex](x-1)^2=4[/tex]

Take the square root of both sides (remember the [itex]\pm[/itex]).

You're right Mentallic but nae99 is doing such silly mistakes here in this thread.

@nae99- take care, since you need to consider this case:- x-1>0
 
  • #34


thanks much
 

Related to 2log(x-1) + logx = logx + log4

1. What is the equation "2log(x-1) + logx = logx + log4" used for?

The equation "2log(x-1) + logx = logx + log4" is used to solve for the value of x in logarithmic equations.

2. How do I solve "2log(x-1) + logx = logx + log4"?

To solve "2log(x-1) + logx = logx + log4", you can use the properties of logarithms to combine the terms on each side of the equation and then solve for x.

3. What are the properties of logarithms used in "2log(x-1) + logx = logx + log4"?

The properties of logarithms used in "2log(x-1) + logx = logx + log4" are the product rule, quotient rule, and power rule.

4. Can I solve "2log(x-1) + logx = logx + log4" without using logarithms?

No, logarithms are necessary to solve "2log(x-1) + logx = logx + log4" since the equation is written in logarithmic form.

5. How can I check my solution for "2log(x-1) + logx = logx + log4"?

You can check your solution for "2log(x-1) + logx = logx + log4" by plugging in the value of x into the original equation and seeing if both sides are equal.

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