h(x) = log (6x+5/7x-4)
Find a formula for the inverse function h^-1(x)
h^-1(x) =
My work so far h(x) = log(6x+5/7x-4)
y=log(6x+5/7x-4)
x=log(6y+5/7y-4)
7xy-4x= log(6y+5)
how would I solve from here. Would I solve for y then do the natural log of both sides or do the natural log first then solve for y?
First, you have to note that since log(z) is defined only for z>0, we must have
x < -4/5 or x > 4/7
In that domain, there is an inverse.
You made an error in your last line, which is where the trouble arises. It should be
x = log (6y+5)/(7y-4)
e^x = (6y+5)/(7y-4)
e^x(7y-4) = 6y+5
7e^x y - 4e^x = 6y+5
(7e^x-6)y = 4e^x+5
y = (4e^x+5)/(7e^x-6)
To find the inverse function of h(x) = log((6x + 5)/(7x - 4)), you need to switch the roles of x and y and solve for y.
Starting from your work, we have:
x = log((6y + 5)/(7y - 4))
To remove the logarithm, we can rewrite the equation in exponential form:
10^x = (6y + 5)/(7y - 4)
Now, let's solve for y.
Multiplying both sides of the equation by (7y - 4) to eliminate the denominator:
(7y - 4) * 10^x = (6y + 5)
Expanding the left side:
7y * 10^x - 4 * 10^x = 6y + 5
Rearranging the equation to isolate the variable y terms:
7y * 10^x - 6y = 4 * 10^x + 5
Combining like terms:
y * (7 * 10^x - 6) = 4 * 10^x + 5
Now, solve for y by dividing both sides of the equation by (7 * 10^x - 6):
y = (4 * 10^x + 5) / (7 * 10^x - 6)
Therefore, the formula for the inverse function h^(-1)(x) would be:
h^(-1)(x) = (4 * 10^x + 5) / (7 * 10^x - 6)