Question

I find my mistake, please help.
The two curves y1 = bx^2 and y2 = ln x "touch" at a point.
Calculate the exact value of b.

I get that
y1 = y2 (1) &
y1' = y2' (2)

I solve for b in (1) and substitute it in (2)
But I end up getting 1/(2 sqrt(e))
The answer is: 1/(2e)

Answer 1

I can't find*

Answer 2

strange, i don't get either of those things
i must be doing something wrong
first one gives
\[bx^2=\ln(x)\]
\[b=\frac{\ln(x)}{x^2}\]
second one gives
\[2bx=\frac{1}{x}\]
\[b=\frac{1}{2x^2}\]

Answer 3

books answer is right though
what did you compute?

Answer 4

if we substitute (1) in (2)\[b(\frac{ 1 }{ 2 })=\frac{ 1 }{ 2 } \ln \frac{ 1 }{ 2b }\]
\[b=\ln 1-\ln 2b\]
\[b=-\ln 2b\]
cant move from here

Answer 5

i get
\[\frac{1}{2x^2}=\frac{\ln(x)}{x^2}\]
\[\ln(x)=\frac{1}{2}\]
\[x=\sqrt{e}\] then find \(b\) by setting
\[b(\sqrt{e})^2=\ln(\sqrt{e})\]
\[eb=\frac{1}{2}\]
\[b=\frac{1}{2e}\]

Answer 6

i made mistake its \[\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 } \ln \frac{ 1 }{ 2b }\]
\[1=\ln \frac{ 1 }{ 2b }\]
\[e^1=1/2b\]
\[b=\frac{ 1 }{ 2e }\]