Answer:
Step-by-step explanation:
<u>Step 1: Set d to 9 and c to 5
</u>
Answer:
Answer:
The solution is (1,2)
Step-by-step explanation:
- 2x - y = - 4
x + 2y = 5
Multiply the second equation by 2 so we can eliminate x
2( x + 2y) = 5*2
2x +4y = 10
Add this to the first equation
- 2x - y = - 4
2x +4y = 10
----------------------
3y = 6
Divide each side by 3
3y/3 = 6/3
y = 2
- 2x - y = - 4
x + 2y = 5
Multiply the first equation by 2 to eliminate y
2(- 2x - y) = - 4*2
-4x -2y = -8
Add this to the second equation
-4x -2y = -8
x + 2y = 5
------------------
-3x = -3
Divide by -3
-3x/-3 = -3/-3
x = 1
The solution is (1,2)
as you already know, to get the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".
![\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=\log_2(x+1)\implies \stackrel{\textit{quick switcheroo}}{\underline{x}=\log_2(\underline{y}+1)}\implies 2^x=2^{\log_2({y}+1)} \\\\\\ 2^x=y+1\implies 2^x-1=\stackrel{f^{-1}(x)}{y} \\\\[-0.35em] ~\dotfill\\\\ 2^2-1=f^{-1}(2)\implies 3=f^{-1}(2)](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%20%5C%5C%5C%5C%20log_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7Bf%28x%29%7D%7By%7D%3D%5Clog_2%28x%2B1%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bquick%20switcheroo%7D%7D%7B%5Cunderline%7Bx%7D%3D%5Clog_2%28%5Cunderline%7By%7D%2B1%29%7D%5Cimplies%202%5Ex%3D2%5E%7B%5Clog_2%28%7By%7D%2B1%29%7D%20%5C%5C%5C%5C%5C%5C%202%5Ex%3Dy%2B1%5Cimplies%202%5Ex-1%3D%5Cstackrel%7Bf%5E%7B-1%7D%28x%29%7D%7By%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%202%5E2-1%3Df%5E%7B-1%7D%282%29%5Cimplies%203%3Df%5E%7B-1%7D%282%29)
Answer:
15
Step-by-step explanation:
15 x 2 = 30
