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Mathematics 6 Online
OpenStudy (anonymous):

Solve without a calculator. 2ln(x+3)-ln(x-1)=3ln2 I know the answer is 1 but I dont know why

OpenStudy (kc_kennylau):

Convert the LHS and RHS respectively into one LN expression each.

OpenStudy (kc_kennylau):

That means combine everything

OpenStudy (kc_kennylau):

Some formulas: \[\Large\log_na+\log_nb=\log_nab\]\[\Large\log_na-\log_nb=\log_n\frac ab\]\[\Large m\log_na=\log_na^m\]

OpenStudy (anonymous):

but this is with ln so would it still be the same?

OpenStudy (kc_kennylau):

\(ln\) is just \(log\) with base \(e\)

OpenStudy (anonymous):

oh yeah. so would \[\frac{ 2\ln(x+3) }{\ln(x+1)} = 3\ln2\] ?

OpenStudy (kc_kennylau):

Nope

OpenStudy (anonymous):

then I dont understand

OpenStudy (kc_kennylau):

You used the formula wrongly

OpenStudy (anonymous):

How do I use it then?

OpenStudy (kc_kennylau):

Use the third formula first

OpenStudy (kc_kennylau):

Remember that in my formulas, when you combine two things together there will be only one "ln" symbol left

OpenStudy (anonymous):

I dont know how to do it separately but is it \[\ln\frac {x+3 }{x+1 }) ^{2}\]

OpenStudy (kc_kennylau):

Let's do it like this, I'll provide you all the steps, and you'll tell me if you don't understand any step

OpenStudy (kc_kennylau):

\[\begin{array}{rcl} 2\ln(x+3)-\ln(x-1)&=&3\ln2\\ \ln(x+3)^2-\ln(x-1)&=&\ln2^3\\ \ln\left[\frac{(x+3)^2}{x-1}\right]&=&\ln8\\ \frac{(x+3)^2}{x-1}&=&8\\ (x+3)^2&=&8(x-1) \end{array}\]

OpenStudy (anonymous):

ohhhhh duhh. i feel dumb now.

OpenStudy (kc_kennylau):

Nah, it's not your fault, it's only that I haven't explained things clearly enough.

OpenStudy (anonymous):

its actually x+1 in my book not x-1 but still how does tht equal 1?

OpenStudy (kc_kennylau):

\[\begin{array}{rcl} 2\ln(x+3)-\ln(x+1)&=&3\ln2\\ \ln(x+3)^2-\ln(x+1)&=&\ln2^3\\ \ln\left[\frac{(x+3)^2}{x+1}\right]&=&\ln8\\ \frac{(x+3)^2}{x+1}&=&8\\ (x+3)^2&=&8(x+1) \end{array}\]

OpenStudy (anonymous):

ya i got 4 and -2 as answers tho

OpenStudy (kc_kennylau):

\[\begin{array}{rcl} (x+3)^2&=&8(x+1)\\ x^2+6x+9&=&8x+8\\ x^2-2x+1&=&0\\ (x-1)^2&=&0\\ x-1&=&0\\ x&=&1 \end{array}\]

OpenStudy (anonymous):

i obviously cant think today.

OpenStudy (kc_kennylau):

Well everyone got their time in a bad mood

OpenStudy (anonymous):

want to help me with a different one?

OpenStudy (kc_kennylau):

Okay, post another question

OpenStudy (anonymous):

and tag you? or u will find it?

OpenStudy (kc_kennylau):

Okay

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