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Mathematics 22 Online
Ballery1:

math hep plz

Ballery1:

solve the given equation for *X* if possible.|dw:1568299284713:dw|

Ballery1:

i need someone to hold my hand and guide me through this problem

Razor:

e^2x ??? Or what? I need some context if this isn't the entire quation

Ballery1:

that *2x* is pretty up there bro... it can only be power to be that high.

Razor:

XD Alrightttt sheeesh

Razor:

@dude I haven't learned this yet

Razor:

Don't complain XD

Ballery1:

k thanks for trying mate. appreciated.

dude:

Okay you first want to get rid of e (since there are no common bases here) So do ln() on both sides Btw do you know what `ln()` is?

Ballery1:

the natural base to natural log is *E* sooo

Ballery1:

i can plug that in ??

dude:

ln is just the opposite of e, so you're basically cancelling it out |dw:1568299915021:dw| Also remember to use the law of logs (Moving the power to the front)

Ballery1:

icy

Ballery1:

|dw:1568300201337:dw|

Ballery1:

omg i just realized something...i've actually already done that question and i corrected my mistake on the paper... hold up, let me post the actual question that i need your help with...sorry

dude:

Its okay dont worry \(\large \frac{ln(4)}{2}\) is also \(ln(2)\)

Ballery1:

but my book says it's this...|dw:1568300430423:dw|

Ballery1:

next question :) |dw:1568300466234:dw|

Ballery1:

|dw:1568301988299:dw|

Ballery1:

|dw:1568302222670:dw|

dude:

Be careful You cannot divide \(\large\frac{ln(6)}{ln(3)}\)

mhchen:

Use this formula: \[\frac{\log_{a}(b)}{\log_{a}(c)} = \log_{c}(b)\]

Ballery1:

so what do i do ?

Ballery1:

|dw:1568306396271:dw|

dude:

I am not sure whether you need to do that but just look at the variables chen used |dw:1568306809864:dw|

Ballery1:

ok? im solving for x so i need to isolate all the variables on one side

dude:

Yeah \(\large \frac{ln(6)}{ln(3)}+1\) is right unless they specifically asked you to simplify completely

Ballery1:

do you know the answer is |dw:1568307258001:dw|

Ballery1:

i've check the answer to that question.. twice.

JSVSL7:

Apply the base changing formula to log3(2)

Ballery1:

|dw:1568307465692:dw|

Ballery1:

gimme a little more info.. plz

JSVSL7:

JSVSL7:

As an example ^

Ballery1:

omg... |dw:1568307641030:dw|

JSVSL7:

Wasn't 2 being added to the logarithmic relation?

JSVSL7:

Leave the 2 alone you can evaluate log3(2) and add

Ballery1:

yes... sorry.. letme correst.

Ballery1:

|dw:1568307763617:dw|

Ballery1:

ok what do i do now ? :/

JSVSL7:

Evaluate using a calculator

Ballery1:

no.... i literally gave you the answer from the textbook... wait, let me rewrite the answer.. |dw:1568307941824:dw|

JSVSL7:

Is it asking you to evaluate or simplify

Ballery1:

but with arithmatically i'm getting this.. |dw:1568307980361:dw|

Ballery1:

the question says *solve for the given equation for x, if possible

JSVSL7:

|dw:1568308084759:dw| Where's the x here? It looks like 2 + log3(2)

Ballery1:

that's logbase 3 and number 2. the whole thing is equal to x

Ballery1:

they didn't write x=............the answer that i posted up there^^ just the answer

mhchen:

|dw:1568309672384:dw|

mhchen:

Keep going

Ballery1:

wait, we started from .... |dw:1568310024903:dw|

Ballery1:

|dw:1568310095965:dw|

Ballery1:

so far so good?

Ballery1:

jus rework with me on this.. i forgot where we were..thanks :)

mhchen:

|dw:1568310280614:dw|

Ballery1:

|dw:1568310259337:dw|

Ballery1:

good so far? or am i making some sort of mistake?

mhchen:

|dw:1568310395373:dw|

Ballery1:

oh ok

Ballery1:

|dw:1568310486215:dw|

Ballery1:

|dw:1568310568003:dw|

Ballery1:

|dw:1568310601768:dw|

Ballery1:

is there anything else left to do ?

mhchen:

|dw:1568310649371:dw|

Ballery1:

but 3 isn't the common base tho. the common base is e

Ballery1:

and how did log turn into natural log?

mhchen:

\[\frac{\ln a}{\ln b} = \log_{b}(a)\]

mhchen:

\[\frac{\log_{e}(6)}{\log_{e}(3)} = \log_{3}(6) = \log_{3}(2*3)=\log_{3}(2)+\log_{3}(3)\]

Ballery1:

icy

mhchen:

|dw:1568310856809:dw|

Ballery1:

icy

Ballery1:

i think i got it... thanks mate. really appreciated your help ^~^

Ballery1:

next question.

Ballery1:

i'll post in the new thread

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