This question can also done by taking log both sides, then applying log(m)^n = nlogm. After this seperating all x on one side and constant on other side. The problem is the constant is a bit weird and complex to handle so take it as "c" or something. Then find the value of x in terms of c and substitute the value of c later on. The value of c is actually (log7/log2) with base 10.
Idk why it bothered me so much that you didn't use the natural log hahahaha. It's been a while for me, but would a natural log answer be easier (or maybe more straightforward) to differentiate/integrate?
So the first solition is (3log(2)+log(7))/(2log(7)-4log(2)) which can be simplified into log(56)/log(49/16). Since both of the logs have the same base we can use the base change formula to get log_49/16(56)
I have a question. So, the first equation could have this resolution? 1/2 . Log(7/4) 56 By the way, the solution to the second equation is Log(3/5)5625 (but can be 2 . Log(3/5)75).
For more examples of how to solve exponential equations from basics to hard, please see ruclips.net/video/K8CQbSD9wis/видео.html
That is about 3.5965 if anyone was wondering
This question can also done by taking log both sides, then applying log(m)^n = nlogm. After this seperating all x on one side and constant on other side. The problem is the constant is a bit weird and complex to handle so take it as "c" or something. Then find the value of x in terms of c and substitute the value of c later on. The value of c is actually (log7/log2) with base 10.
Idk why it bothered me so much that you didn't use the natural log hahahaha. It's been a while for me, but would a natural log answer be easier (or maybe more straightforward) to differentiate/integrate?
Log 49/15 (56) is crazy bro
16
no 3.6 ish
so this and that cancel
Now It's time to show that the two solutions are actuallt the same
So the first solition is (3log(2)+log(7))/(2log(7)-4log(2)) which can be simplified into log(56)/log(49/16). Since both of the logs have the same base we can use the base change formula to get log_49/16(56)
3^x/3^2 = 5^x . 5^4
3^x/9 = 5^x . 625
3^x = 5^x . 5625
3^x/5^x = 5625
(3/5)^x = 5625
x = log 3/5 (5625)
I have a question. So, the first equation could have this resolution? 1/2 . Log(7/4) 56
By the way, the solution to the second equation is Log(3/5)5625 (but can be 2 . Log(3/5)75).
It's always a pleasure to learn thank to you.
log3/5 (5625) ?
Yes
since 5625 is 75^2, you can further simplify into
= log 3/5 (75^2)
= 2 * log 3/5 (75)
X.ln (49/16)= ln(56)
X= ln(56)/ln(49/16)
Why do you want to change the base?
Luvley jubbley 👍👍👏👏
Amazing ❤
Answer for the last equation is log_3/5(625/9) or approx -8.301
the 625/9 is actually a 625*9
great video
i'm not too good at exponents.
maybe x = log5(5625) ?
❤❤❤
interesting how I watched the originally vid this morning
there should be a way to solve this using e
There is a way using natural logs(I.e ‘ln’)