- #1

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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)

- Thread starter nae99
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- #1

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

log(x-1)^2 + logx = logx (4)

- #2

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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

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i dont get it

- #4

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so are u saying it should be:

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.

log (2x - x)^2 = 4x

- #5

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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

- 3,812

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Noooooo, :)so are u saying it should be:

log (2x - x)^2 = 4x

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

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log (x-1) = log4

i dont think i understand that

- #8

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2log (x-1) - log 4 = 0Noooooo, :)

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]

- #9

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There you go! I'd have kept log(4) on the right side and then just solve for x.2log (x-1) - log 4 = 0

- #10

- 3,812

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Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave.2log (x-1) - log 4 = 0

- #11

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log (x-1)^2 - log 4 = 0Yep, that's what i meant to say. Now try converting 2log (x-1) using the identities i gave.

- #12

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- #13

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- #14

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It would belog (x^2-1) /4 = 0

[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.

- #15

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ok so i would do this next but i dont think its rightIt 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.

(x^2-1) /4 = 1

- #16

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I don't understand how you convert (x-1)ok so i would do this next but i dont think its right

(x^2-1) /4 = 1

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

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ok, but i thought (x-1) meant (x-1) (x-1)

- #18

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I still didn't get youok, but i thought (x-1) meant (x-1) (x-1)

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

- #19

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so i would end up with:I still didn't get you

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

x^2-2x+1 = 4

- #20

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

- #21

- 3,812

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Yes, now proceeding further you getso i would end up with:

x^2-2x+1 = 4

[tex]x^2-2x-3=0[/tex]

Now this is quadratic equation, try to solve it. :)

- #22

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x = -(-2) [itex]\pm[/itex] [itex]\sqrt{}[/itex] (-2)^2 - 4*1*(-3)[itex]/[/itex] 2*1Yes, 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] -4+12[itex]/[/itex] 2

how is that

- #23

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What is the square of (-2)?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

- #24

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i dont know

- #25

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