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

[nae99]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[Omega_Prime]

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]

i don't get it

 

[nae99]

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

 

[Omega_Prime]

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.

 

[Saitama]

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 $\log x$ on both sides.
Then shift $\log 4$ to the left side remaining 0 at the right side.
Now using these two identities, try to figure out the answer:-
$$p(\log_a b)=log_a b^p$$
and
$$\log_a u-\log_a v=\log_a \frac{u}{v}$$

 

[nae99]

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

 

[nae99]

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

2log (x-1) - log 4 = 0

 

[Omega_Prime]

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.

 

[Saitama]

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.

 

[nae99]

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

log (x-1)^2 - log 4 = 0

 

[Saitama]

log (x-1)^2 - log 4 = 0

Now use the second identity.

 

[nae99]

Now use the second identity.

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

 

[Saitama]

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

It would be
$$\log \frac{(x-1)^2}{4}=0$$

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

Now solve it further.

 

[nae99]

It would be
$$\log \frac{(x-1)^2}{4}=0$$

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

Now solve it further.

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

(x^2-1) /4 = 1

 

[Saitama]

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)?
That's wrong!

Solving this equation:-
$$\frac {(x-1)^2}{4}=1$$
we get,
$$(x-1)^2=4$$
Now try to do it.

 

[nae99]

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

 

[Saitama]

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

I still didn't get you
$$(x-1)^2=(x-1)(x-1)=x^2-2x+1$$

 

[nae99]

I still didn't get you
$$(x-1)^2=(x-1)(x-1)=x^2-2x+1$$

so i would end up with:

x^2-2x+1 = 4

 

[nae99]

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

 

[Saitama]

so i would end up with:

x^2-2x+1 = 4

Yes, now proceeding further you get
$$x^2-2x-3=0$$
Now this is quadratic equation, try to solve it. :)

 

[nae99]

Yes, now proceeding further you get
$$x^2-2x-3=0$$
Now this is quadratic equation, try to solve it. :)

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

x= -(-2) $\pm$ $\sqrt{}$ -4+12$/$ 2

how is that

 

[Saitama]

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

x= -(-2) $\pm$ $\sqrt{}$ -4+12$/$ 2

how is that

What is the square of (-2)?

 

[nae99]

i don't know

 

[nae99]

What is the square of (-2)?

is it -4

 

[Saitama]

is it -4

Nooo, it's 4.
(-2)²=(-2)(-2)=4

 

[nae99]

oh ok

 

[Saitama]

oh ok

Now what's the answer?

 

[nae99]

Now what's the answer?

x= -(-2) ± √ 4+12/ 2

x = -(2) $\pm$$\sqrt{}$ 16 $/$ 2

x = -(-2) $\pm$ 4 $/$ 2

x = -(-2) + 4 $/$ 2

x = 6$/$ 2

x= 4
OR

x = -(-2) - 2 $/$ 2

x = -2$/$ 2

x = -1

 

[Saitama]

x= -(-2) ± √ 4+12/ 2

x = -(2) $\pm$$\sqrt{}$ 16 $/$ 2

x = -(-2) $\pm$ 4 $/$ 2

x = -(-2) + 4 $/$ 2

x = 6$/$ 2

x= 4
OR

x = -(-2) - 2 $/$ 2

x = -2$/$ 2

x = -1

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

 

[nae99]

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

 

[Mentallic]

That's too much unnecessary work. From,

$$(x-1)^2=4$$

Take the square root of both sides (remember the $\pm$).

 

[Saitama]

That's too much unnecessary work. From,

$$(x-1)^2=4$$

Take the square root of both sides (remember the $\pm$).

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

 

[nae99]

thanks much