for supercrazy:
how to solve $$3^{4x-7}=4^{2x+3}$$

Question

for supercrazy:
how to solve $$3^{4x-7}=4^{2x+3}$$

myininaya · Answer

Do you remember $$\ln(x^r)=r \ln(x)$$

myininaya · Answer

i got stood up

anonymous · Answer

lol

amistre64 · Answer

those loger-epo rules make more sense when defined by calculus

anonymous · Answer

@Myini-Did I do something? o.o Sorry I will delete my comment.

myininaya · Answer

No this person wanted to know how to do this problem in chat and I said was going to post it here and she never came.  :(

anonymous · Answer

@Myini-You did your work nevertheless, shows you're a sincere person :)

anonymous · Answer

Just to practice latex $$\begin{align}3^{4x-7}=4^{2x+3} &<=> (4x-7)ln(3) &&=(2x+3)ln(4) \ &<=> x4ln(3)-7ln(3) &&= x2ln(4)+3ln(4) \ &<=> x(4ln(3)-2ln(4)) &&= 3ln(4)+7ln(3) \ & <=> x &&={ 3ln(4)+7ln(3) \over 4ln(3)-2ln(4)} \ or\ & x={ln(4^3.3^7) \over ln(3^4.4^2) }&& = {ln(139968)\over ln(1296)}\approx 1.65\end{align} $$

myininaya · Answer

Pretty mkone!

myininaya · Answer

well i think i see something just a little off on your solution

myininaya · Answer

$$4 \ln(3)-2 \ln(4)=\ln(3^4)-\ln(4^2)=\ln(\frac{3^4}{4^2})$$

anonymous · Answer

I'm not getting what's off is it the presentation. If so I agree $ln({3^4 \over 4^2}) $ looks better but then what about $2ln({3^2 \over 4}) $ ?