![\bf log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad % exponential notation 2nd form {{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y\\ \\ \quad \\ thus\\ ----------------------------\\ 8^5=32768\qquad \iff\qquad log_8(32768)=5](https://tex.z-dn.net/?f=%5Cbf%20%0Alog_%7B%7B%20%20a%7D%7D%7B%7B%20%20b%7D%7D%3Dy%20%5Ciff%20%7B%7B%20%20a%7D%7D%5Ey%3D%7B%7B%20%20b%7D%7D%5Cqquad%5Cqquad%20%0A%25%20%20exponential%20notation%202nd%20form%0A%7B%7B%20%20a%7D%7D%5Ey%3D%7B%7B%20%20b%7D%7D%5Ciff%20log_%7B%7B%20%20a%7D%7D%7B%7B%20%20b%7D%7D%3Dy%5C%5C%0A%5C%5C%20%5Cquad%20%5C%5C%0Athus%5C%5C%0A----------------------------%5C%5C%0A8%5E5%3D32768%5Cqquad%20%5Ciff%5Cqquad%20log_8%2832768%29%3D5)
so... notice the example there, with 8
so.. you tell us, what is the relationship then?
Given:
The given expression is
![-4(x+2)-2x+4](https://tex.z-dn.net/?f=-4%28x%2B2%29-2x%2B4)
To find:
The two other equivalent expressions.
Solution:
We have,
![-4(x+2)-2x+4](https://tex.z-dn.net/?f=-4%28x%2B2%29-2x%2B4)
Using distributive property, we get
![=-4(x)-4(2)-2x+4](https://tex.z-dn.net/?f=%3D-4%28x%29-4%282%29-2x%2B4)
![=-4x-8-2x+4](https://tex.z-dn.net/?f=%3D-4x-8-2x%2B4)
So, one equivalent expression is
.
On further simplification, we get
![=(-4x-2x)+(-8+4)](https://tex.z-dn.net/?f=%3D%28-4x-2x%29%2B%28-8%2B4%29)
![=6x-4](https://tex.z-dn.net/?f=%3D6x-4)
Therefore, the second and fully simplified equivalent expression is
.
Solve for the first variable in one of the equations, then substitute the result into the other equation.
3. (1/2, -1,2)
4. (2, -4)
You can get step by step at:
https://www.mathpapa.com/algebra-calculator.html
Answer:
g(a+1)=5a+2
Step-by-step explanation:
so (a+1) is the value of x
you can write it like this:
x=a+1
using substitution, we can swap out all 'x's for 'a+1's. like so;
g(a+1)=5(a+1)-3 -- given
=5a+5-3 -- distribute
=5a+2 -- simplify!
Hope i helped! have a great day! please mark brainliest!!
Evaluate this...
![\frac{x^2+3x-4}{x-1}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%5E2%2B3x-4%7D%7Bx-1%7D%20)
the x at the bottom cancels out one of the xs at the top so it will be
![\frac{x^2+2x-4}{-1}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%5E2%2B2x-4%7D%7B-1%7D%20)
Now the one cancels out one of the 1s in the -4 so
![{x^2+2x-3}](https://tex.z-dn.net/?f=%7Bx%5E2%2B2x-3%7D)
Hope this helps :)