Question

Solve each equation below

Answer 1

\[\log (\frac{ x }{ 10 })=-2+2\log x\]

Answer 2

\[\log (\frac{ x }{ 10 })=-2+2\log x\]
\[\ e^{\log (\frac{ x }{ 10 })}=e^{-2+2\log x}\]
\[(\frac{ x }{ 10 })=e^{-2}*e^{2\log x}\]
\[(\frac{ x }{ 10 })=e^{-2}*e^{\log x^{2}}\]
\[(\frac{ x }{ 10 })=e^{-2}*x^{2}\]
\[e^{-2}*x^{2}-\frac {x} {10} =0\]
\[x(e^{-2}*x-\frac{1}{10})=0\]
\[x=0\] or
\[x= \frac{e^{2}}{10}\]

Answer 3

e = 10

Answer 4

wow...okay

Answer 5

It looks so simple once it's all solved but i cant ever quite get it...

Answer 6

if there's a log you have to raise both sides by the base.

Answer 7

no matter what?

Answer 8

well, if you have log = log then you can just take the log out.
for example: log (5x+3)=log (2), goes to 5x+3 = 2

Answer 9

okay that makes sense.

Mind helping with the next one?

Answer 10

sure

Answer 11

*phew* okay!
\[\sqrt{3x+6}-x-2=0\]

Answer 12

I feel like its pretty straight forward and I come close to the answer but its never quite right. this is what I did.

Answer 13

And i just accidentally deleted everything...

Answer 14

but i ended up with 0=x^2+x-2

Answer 15

\[\sqrt{3x+6}-x-2=0\]
\[\sqrt{3x+6}=x+2\]
\[(\sqrt{3x+6})^{2}=(x+2)^{2}\]
\[{3x+6}=(x+2)*(x+2)\]
\[3x+6=x^{2}+4x+4\]
\[x^{2} + x - 2 = 0\]
\[(x+2)(x-1)=0\]
\[x= -2, x=1\] but now you have to check your answers
\[\sqrt{3(-2)+6}-(-2)-2=0\]
\[\sqrt{-6+6}+2-2=0\]
\[\sqrt{0}+0=0\] ture, x=-2 is an answer,
\[\sqrt{3(1)+6}-(1)-2=0\]
\[\sqrt{9}-3=0\]
\[3-3=0\] ture, x=1 is answer,
therefore: x= 1 and x= -2

Answer 16

So i was on the right track i just missed that last step...darn. So close....

Answer 17

okay another seemingly straightforward one...\[\sqrt{t+7}-\sqrt{t}=1\]

Answer 18

When I look at it the first thing i do is \[(\sqrt{t+7}-\sqrt{t}=1)^{2}\]

Answer 19

but if you do that you end up with 7=1 which isnt right of course.

Answer 20

more the sqrt (t) to the other side before you square both side

Answer 21

thank you so much...Like i said so close but so far away...

Answer 22

\[t+7 = t+ 2\sqrt{t} +1\]
\[3 = \sqrt{t}\]
t= 9 then go back and check it

Answer 23

wait, where did the t+2tsqrd come from?

Answer 24

\[(\sqrt{t+7}=1+\sqrt{t})^{2}\]
\[t+7 = (1+\sqrt{t})^{2}\]
\[t+7 = t+ 2\sqrt{t} +1\]

Answer 25

ohhhh okay so then you subtract t and 1, then divide both sides by 2 then boom. 9.

Answer 26

ya, i was doing it quickly,

Answer 27

i got to get to class now. good luck.

Answer 28

thanks so much.