Woahhh! you are going to need some fancy work here
let's look at y = x^4 vs y = 3x/4 + 7
https://www.wolframalpha.com/input/?i=plot+y+%3D+4%5Ex+,+y+%3D+3x%2F4+%2B+7
looks like we have 2 solutions, one between -9 and -10 and another around 1.5
Do you know Calculus, thus Newton's Method?
Find x
(4^x)=((3/4)x)+7
Thank you!
2 answers
I suspect a typo. However, we can solve
4^x=3/4^x+7
Now, let u = 4^x and we have
u^2 - 7u - 3 = 0
u = (7±√61)/2
We want only the real solution:
u = (7+√61)/2
so, x = log4 (7+√61)/2
or
x = (log (7+√61)/2)/log4
4^x=3/4^x+7
Now, let u = 4^x and we have
u^2 - 7u - 3 = 0
u = (7±√61)/2
We want only the real solution:
u = (7+√61)/2
so, x = log4 (7+√61)/2
or
x = (log (7+√61)/2)/log4