Question

Ok let me make a easy question FIND d^2y/dx^2 if y=log(x^2/e^x)

Answer 1

First time you've used parentheses!

Except we don't know what base the log is.

Answer 2

\[\log x^{2}\div e^{x}\]

Answer 3

Now it's even less clear than it was.

Answer 4

\[\frac{ x^{2} }{ e^{x} }\]

Answer 5

Is it log base 10 I guess??

Answer 6

\[\log \frac{ x ^{2} }{ e ^{x} }\]

Answer 7

Thats super clear

Answer 8

So it's log base 10, correct? Then you may want to use the change of base formula first, get it to base e.

Answer 9

IDK

Answer 10

thats why i am asking

Answer 11

Uhhh if you don't know if it's log base 10, how am i supposed to?

Answer 12

It is not log base 10 lol use log formula

Answer 13

Then what base is it?!

Answer 14

log/log

Answer 15

Lol you can that eqn like this too

Answer 16

\[\log x^{2}-e^{x}\]

Answer 17

standard log formulas

Answer 18

Yeah, so i guess we can assume it's log base 10... so change the base to e, you should know how to do that

Answer 19

\[\large y = \frac{ \ln x^2 }{ \ln 10 }-\frac{ \ln e^x }{ \ln 10 }\] \[\large y = \frac{ 2\ln x }{ \ln 10 }-\frac{ x }{ \ln 10 }\]

Answer 20

Assuming natural logarithm (most common anyway)
\[y = \ln (\frac{ x^2 }{ e^x }) = \ln (x^2) - \ln(e^x) = \ln(x^2) - x\ = 2 \ln(x) - x\]
\[\frac{ dy }{ dx } = \frac{ 2 }{ x } - 1\]
\[\frac{ d^2y }{ dx^2 } = -\frac{ 2 }{ x^2 }\]