Solve the equation $$log_4(x)+log_7(x)=1$$

Question

turingtest · Answer

@purplec16 could you please not post all in caps? It makes it seem as though you are yelling, and is a bit harsh on the eyes.
thanks

turingtest · Answer

also the fastest way to get help is to just type the actual question in the post, so you should probably go ahead and do that

anonymous · Answer

oh... I wasn't yelling... sorry I forgot above that...

turingtest · Answer

it's cool, a lot of people do it

anonymous · Answer

I've had questions up for hours and noone helped me...

anonymous · Answer

Do you know how to solve $$log_4(x)+log_7(x)=1$$

turingtest · Answer

I think I have a plan... let me work it out

anonymous · Answer

Yay! :D

turingtest · Answer

do you know the change of base log formula?

anonymous · Answer

I know it... but my lecturer skipped it... and I don't know how to use it...

turingtest · Answer

@Callisto i have a feeling that you have a better idea than mine, so feel free to put your solution if you have one

turingtest · Answer

I was doing something like the change of base formula, which can be written as the change of log base a to log base b is$$\log_bx={\log_b x\over \log_b a}$$

turingtest · Answer

so taking b=e we can make this a natural logarithm$${\ln x\over\ln4}+{\ln x\over\ln7}=1$$$$\ln7\cdot\ln x+\ln4\cdot\ln x=\ln4\cdot\ln7$$$$\ln x={\ln4\cdot\ln7\over\ln7+\ln4}$$

anonymous · Answer

how'd you go from the first to second step?

turingtest · Answer

factor out lnx$$\ln7\cdot\ln x+\ln4\cdot\ln x=\ln4\cdot\ln7$$$$\ln x(\ln7+\ln4)=\ln4\cdot\ln7$$$$\ln x={\ln4\cdot\ln7\over\ln7+\ln4}$$

turingtest · Answer

oh first to second?
the change of base formula I wrote above

turingtest · Answer

$$\log_bx={\log_b x\over \log_b a}$$

callisto · Answer

$${\ln x\over\ln4}+{\ln x\over\ln7}=1$$$${\ln7\ln x\over\ln4 \ln7}+{\ln4\ln x\over\ln4\ln7}=1$$$${\ln7\ln x+\ln4\ln x\over\ln4\ln7}=1$$$$\ln7\ln x+\ln4\ln x=\ln4\ln7$$

turingtest · Answer

oh, that step :P
I wasn't sure which one you were talking about I guess

anonymous · Answer

Yeah and it's the same thing @Callisto... what did you do for your second line and why?

turingtest · Answer

She just got a common denominator between the two fractions

callisto · Answer

Addition of fraction... you need to have the same denominator for that, right?

anonymous · Answer

Yes... so you multiply each... but I am not seeing it when you're skipping steps.. or idk... it's just a little more difficult with logarithm, sigh... I feel so confused and defeated

callisto · Answer

Multiplying the same thing in numerator and denominator, adding them with the same denominator, multiplying the denominator to both sides.... Which step did I skip? :O

Is it clearer to you now??

anonymous · Answer

actually no... isn't the common denominator ln4ln7...

callisto · Answer

Which part you don't understand?

anonymous · Answer

Oh, I just realized the last line is what I was implying to but I didn't understand your method but that's okay... as of now I don't understand how you got the first line and how to solve the last line...

callisto · Answer

If you don't mind... how do you simplify it? $$\frac{1}{b}+\frac{1}{a}$$

anonymous · Answer

$$\frac{a+b}{ba}$$

I got what you did with the fraction.. but I don't know how to solve the log and I got the first line now too :D

callisto · Answer

''solve the log''? 
If you have lnx = 1, how do you solve it?

anonymous · Answer

e^1=x

anonymous · Answer

I'm really sorry but I haven't done any of the complicated with lnx*ln7 I don't know what to do with that... does is become ln 7x?

callisto · Answer

No... lnx * ln7 ≠ ln (7x)

anonymous · Answer

I don't know how to add two log's...

callisto · Answer

ln7 + ln 4 = ln (7x4) = ...?

anonymous · Answer

28

anonymous · Answer

but... but... :\... why are log's so confusing I can't feel the right side of my head...

callisto · Answer

I don't know too :|

anonymous · Answer

So it's lnx^2ln(28)=ln4ln7

callisto · Answer

No.... you should ''isolate'' x instead.

anonymous · Answer

by factoring?

callisto · Answer

No. Like the way you solve lnx =1
Now, it's $\large lnx = \frac{ln4ln7}{ln4+ln7}$

anonymous · Answer

Doesn't that involving factoring the lnx from the left hand side...
$$lnx(ln7+ln4)=ln4ln7$$
maybe I said it wrong... Idk...

anonymous · Answer

So this $$x=e^{\frac{ln4ln7}{ln28}}$$ In the exponent is...$$\frac{ln4ln7}{ln28}$$

anonymous · Answer

$$x \approx 2.246908864...$$

callisto · Answer

Yup!

anonymous · Answer

Sigh... that was rough... sorry for "yelling" earlier @TuringTest but... I'm preparing for finals and I'm really anxious, stressed, depressed/emotional... and really tired. BUT...Thanks so much @Callisto and @TuringTest  I really really really do appreciate your help! Like seriously, you all are awesome!!! I'm so glad I've found you... lol :D.

callisto · Answer

'' I'm really anxious, stressed, depressed/emotional.'' => That's me!!!!

No worries! Everything will be fine!!! Good luck for the finals too!

anonymous · Answer

Thanks so much! Like I really appreciate you all... because I would have been a nervous wreck if I didn't learn this...and might have forgot how to add/divide/subtract/multiply... lol!!!

anonymous · Answer

Would you mind helping me very quickly with another one, in another post... it shouldn't be as long because I think I grasped the basic idea...

callisto · Answer

Just post the question. I'm sure there's always someone to help!

anonymous · Answer

Okay :D, thanks