- #1
nae99
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Homework Statement
2log(x-1) + logx = logx + log4
Homework Equations
The Attempt at a Solution
log(x-1)^2 + logx = logx (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.
nae99 said:so are u saying it should be:
log (2x - x)^2 = 4x
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.
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]
nae99 said:2log (x-1) - log 4 = 0
nae99 said:2log (x-1) - log 4 = 0
Pranav-Arora said:Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave.
nae99 said:log (x-1)^2 - log 4 = 0
Pranav-Arora said:Now use the second identity.
nae99 said:log (x^2-1) /4 = 0
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.
nae99 said:ok so i would do this next but i don't think its right
(x^2-1) /4 = 1
nae99 said:ok, but i thought (x-1) meant (x-1) (x-1)
Pranav-Arora said:I still didn't get you
[tex](x-1)^2=(x-1)(x-1)=x^2-2x+1[/tex]
nae99 said:so i would end up with:
x^2-2x+1 = 4
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. :)
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
Pranav-Arora said:What is the square of (-2)?
nae99 said:is it -4
nae99 said:oh ok
Pranav-Arora said:Now what's the answer?
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
Pranav-Arora said:What are you doing?
6/2=3 and 2-2=0.
Your both the answers are wrong. Correct them.
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]).
The equation "2log(x-1) + logx = logx + log4" is used to solve for the value of x in logarithmic equations.
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.
The properties of logarithms used in "2log(x-1) + logx = logx + log4" are the product rule, quotient rule, and power rule.
No, logarithms are necessary to solve "2log(x-1) + logx = logx + log4" since the equation is written in logarithmic form.
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.