exponentials help Watch

panjabiflower
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#1
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#1
if the curves y=ln (4x-3) and y= 1+ ln x cross x axis at a and b what are the coordinates of a and b
also if the two curves intersect at c how would you be able to find out the x coordinate of c in terms of e?

thanx to all the people who reply
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Pork and Beans
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A line crosses the x axis at y=0, so solve ln(4x-3)=0 and 1+ln x=0.

To find the intersection, solve ln(4x-3)=1+ln x.
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generalebriety
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Well, y = ln(4x - 3) crosses the x-axis when y = 0, so solve 0 = ln(4x - 3). (Remember that 0 = ln 1.) Also solve 0 = the right hand side of your other equation (which I think you've mistyped).

When the two curves intersect, just set their right hand sides equal and solve the equation for x.
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panjabiflower
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how would you solve
ln (4x-3)=0 and
0=1+lnx
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Pork and Beans
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(Original post by panjabiflower)
how would you solve
ln (4x-3)=0 and
0=1+lnx
ln(4x-3) = 0 \Rightarrow 4x - 3 = e^0 = 1

The second one is simple.
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panjabiflower
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#6
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plz tell me how to work out the 2nd one
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panjabiflower
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#7
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is the answer 0.37
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panjabiflower
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#8
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how do you work out the x coordinate of c?
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Pork and Beans
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(Original post by panjabiflower)
is the answer 0.37
Yes, although you should probably leave it as e^{-1} or \frac{1}{e}

To find the x coordinate of c, you need to solve ln(4x-3)=1+ln x.

Spoiler:
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ln(4x-3)=1+ln x \Rightarrow ln(4x-3)-ln x=1 \Rightarrow ln(\frac{4x-3}{x}) = 1
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panjabiflower
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hi is the x value for c 2.31?
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Pork and Beans
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(Original post by panjabiflower)
hi is the x value for c 2.31?
Close, but you have to leave it in terms of e.
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panjabiflower
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i worked it out again and realised i made a mistake answer is 2.34 plz tell me how to leave answer in terms of e plz
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(Original post by panjabiflower)
i worked it out again and realised i made a mistake answer is 2.34 plz tell me how to leave answer in terms of e plz
Well you should have it in terms of e in your working unless you've been using trial and improvement.
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panjabiflower
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#14
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this is how i worked it out:
ln (4x-3)-lnx=1
ln (4x-3/x)=1
lna=c
e^c=a
e^1=2.72=a
4x-3/x=2.72
4x-3=2.72x
1.28x=3
x=3/1.3
=2.32 (2.dp)
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Pork and Beans
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(Original post by panjabiflower)
this is how i worked it out:
ln (4x-3)-lnx=1
ln (4x-3/x)=1
lna=c
e^c=a
e^1=2.72=a
4x-3/x=2.72
4x-3=2.72x
1.28x=3
x=3/1.3
=2.32 (2.dp)
Use e in your working instead of 2.72
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panjabiflower
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#16
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e=4-3/x
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generalebriety
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#17
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(Original post by panjabiflower)
this is how i worked it out:
ln (4x-3)-lnx=1
ln (4x-3/x)=1
lna=c
e^c=a
e^1=2.72=a
4x-3/x=2.72
4x-3=2.72x
1.28x=3
x=3/1.3
=2.32 (2.dp)
Just don't change it to 2.72. Leave it in terms of e.
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Pork and Beans
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#18
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#18
(Original post by panjabiflower)
e=4-3/x
Close.

Spoiler:
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\frac{4x-3}{x} = e \newline

4x - 3 = ex \newline

4x - ex = 3 \newline

(4-e)x = 3 \newline

x = \frac {3}{4-e}
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panjabiflower
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#19
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#19
thanx for all the replies i think i now understand
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