Since you know the value of "x", you can plug in the value for "x" in the equation.
[When an exponent is negative, you move it to the other side of the fraction to make the exponent positive.]
For example:
or
![(x^(-2))/(1) =(1)/(x^2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/p67nw5kawtju6a0xid8l9xrpkeqhy4o6vg.png)
or y³
x = -2
f(x) = 9x + 7
f(-2) = 9(-2) + 7 = -18 + 7 = -11
![g(x)=5^x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gjkeg3bos80rcat4iuti5ir8q7txtpzgk5.png)
(idk if you should have it as a decimal or a fraction)
x = -1
f(x) = 9x + 7
f(-1) = 9(-1) + 7 = -9 + 7 = -2
![g(x)=5^x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gjkeg3bos80rcat4iuti5ir8q7txtpzgk5.png)
![g(-1)=5^(-1)=(1)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gbsgjcvts41dxdtmu1ve797ky5bof78iox.png)
x = 0
f(x) = 9x + 7
f(0) = 9(0) + 7 = 7
![g(x)=5^x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gjkeg3bos80rcat4iuti5ir8q7txtpzgk5.png)
![g(0)=5^0=1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ns7the5lxhp25cu8bgz4w3nq0s2abicm76.png)
x = 1
f(x) = 9x + 7
f(1) = 9(1) + 7 = 9 + 7 = 16
![g(x)=5^x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gjkeg3bos80rcat4iuti5ir8q7txtpzgk5.png)
![g(1)=5^1=5](https://img.qammunity.org/2019/formulas/mathematics/middle-school/fonsgqy7a6ktxqpiivdfbpw1z3w0t9hina.png)
x = 2
f(x) = 9x + 7
f(2) = 9(2) + 7 = 18 + 7 = 25
![g(x)=5^x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gjkeg3bos80rcat4iuti5ir8q7txtpzgk5.png)
![g(2)=5^2=25](https://img.qammunity.org/2019/formulas/mathematics/middle-school/1fay6gy9vf9jj35ywbv0bnloeiw1gfj1e4.png)
You need to determine the solution of f(x) = g(x)
Since you know f(x) = 9x + 7 and
, you can plug in (9x + 7) for f(x), and (
) into g(x)
f(x) = g(x)
You can plug in each value of x into the equation
Your answer is x = 2 because when you plug in 2 for x in the equation, you get 25 = 25