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

## Homework Statement

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

## The Attempt at a Solution

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

Related Precalculus Mathematics Homework Help News on Phys.org

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.

i dont get it

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

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.

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
$$\log_a u-\log_a v=\log_a \frac{u}{v}$$

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 dont think i understand that

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

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.

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. 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

log (x-1)^2 - log 4 = 0
Now use the second identity. Now use the second identity. log (x^2-1) /4 = 0

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. 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 dont think its right

(x^2-1) /4 = 1

ok so i would do this next but i dont 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.

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

ok, but i thought (x-1) meant (x-1) (x-1)
I still didn't get you $$(x-1)^2=(x-1)(x-1)=x^2-2x+1$$

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

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

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. :)

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

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)?  i dont know

What is the square of (-2)? is it -4