Solving exponential equations with different bases

[@skinnyladd]

for the second equation I did, (2^4x) x 2 = 3^x (16^x) x 2 = 3^x (3^x)/(16^x) = 2 (3/16)^x = 2 Taking log base (3/16) on both sides x = log base 3/16 (2) x = -0.414 (which is the same as yours) great problems, looking forward to more!

[@l.w.paradis2108]

I like how this was explained so clearly and simply. Crisp, without needless complications.

[@quantumnoctemus]

He teaches better in one second than many teachers in 1 hour.

[@inmu5529]

2:00 his humor is crazy

[@balduran2003]

It's cool how the answer to the second one still has 1,2,3,4 in it.

[@neilgerace355]

"Surprising, right?" Hahahaha

[@sanjaybhowmick4905]

I am highly impressed sir

[@maroonjacketdkid08]

so that's when "log that's not base e" is important 😮

[@charlescox290]

I'm some of my computer science classes, lg was used as a convention for log base 2.

[@tryphonkorm]

Smooth solution yo yhe second equation.

[@Budgeman83030]

I’m beginning to see that you can do a lot using log, ln, and e. Granted it only took forty years after high school to see this

[@weggquiz]

Well explained

[@LythMusic]

(3x+1)log2 = xlog4 3xlog2 + log2 = xlog4 3xlog2 - xlog4 = -log2 x(3log2-log4) = -log2 x = -log2/3log2-log4

[@leonardobarrera2816]

That is a good one!!!

[@thomassidoti5496]

I love this guy. It's just funny because if he would have just took the ln of 2 and 3 in the first place he wouldn't have to re-write the answer

[@NapasimisOrdibarata]

It also the x is ²log (3/16)

[@zurinakasim2918]

(4x+1)ln2=xln3 4xln2+ln2=xln3 xln16+ln2=xln3 ln2=x(ln3-ln16) x=ln2/(ln3-ln16)

[@geopediashorts]

i just used the property of logarithms, where you put the exponent in front of the logarithm

[@gruinfield1129]

I did a more tedious method and got log2/log3-log2(4) which is still correct

[@TheNerdess]

Im supposed to solve using only natural log. that has been my problem finding examples of people solving with that and not log! Stuff like 2^(5x+4)=3^(3x-2) can you just write ln instead of log?

[@carultch]

Since we have the change-of-base rule, it is arbitrary whether you write ln(8)/ln(2) or log(8)/log(2). Both produce the same result. You can solve any problem involving logs, using either natural log or log base ten. Or even some completely different base like log base 2. To do the problem you provided using natural log: Given: 2^(5*x+4) = 3^(3*x - 2) Take the natural log of both sides: ln(2^(5*x + 4)) = ln(3^(3*x - 2)) Use the log property, ln(a^b) = b*ln(a) to pull the exponents out in front: (5*x + 4)*ln(2) = (3*x - 2)*ln(3) Expand, move constants to the right, and variables to the left: 5*x*ln(2) + 4*ln(2) = 3*x*ln(3) - 2*ln(3) 5*x*ln(2) - 3*x*ln(3) = -2*ln(3) - 4*ln(2) Factor the left: [5*ln(2) - 3*ln(3)]*x = -2*ln(3) - 4*ln(2) Isolate x: x=[-2*ln(3) - 4*ln(2)]/[5*ln(2) - 3*ln(3)] This can simplify to: x = -ln(144)/ln(32/27), which evaluates to about -29.25

[@sebastianjohansen2142]

crazy guy

[@ptrakoo5363]

What is 2 log base2 of 3

[@xvoidx_yt1723]

3

[@neilkarlosnatividad4892]

I shouted NO ******* WAY for the second equation

[@boredafmetoo7467]

Nice Shoes and nice video also

[@DavidOlukoga]

Those nikes bruh

[@graemehumfrey3955]

Beautiful

[@mccauleybacalla2228]

wow

[@comeoffmychannel]

Jesus loves yall and God Bless