Find x
(4^x)=((3/4)x)+7
Thank you!
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?
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
To find the value of x in the equation (4^x)=((3/4)x)+7, we can start by isolating the exponential term on one side of the equation.
Let's first simplify the equation by multiplying both sides by 4 to clear the fraction:
4 * (4^x) = 4 * ((3/4)x + 7)
Now, using the property of exponents, we can rewrite the left side as:
(4^(x+1)) = 3x + 28
Next, let's rewrite 4 as a base of 2 raised to the power of 2:
((2^2)^(x+1) = 3x + 28)
Using the exponent rule (a^m)^n = a^(m*n), we can simplify it further:
(2^(2*(x+1))) = 3x + 28
Now, we can distribute the exponent 2*(x+1):
(2^(2x+2)) = 3x + 28
Since both sides have the same base (2), we can equate the exponents:
2x+2 = 3x + 28
By subtracting 2x from both sides, we have:
2 = x + 28
By subtracting 28 from both sides, we get:
-26 = x
Therefore, the value of x in the given equation is -26.