SOLUTION
We want to solve the question below
write an exponential equation in the form y = ab* whose graph passes through points (-3, 24) and (-2, 12).
Now an exponential equation is written in the form of
![y=a(b)^x](https://img.qammunity.org/2023/formulas/mathematics/college/x4kto8751eypenmr0i3prbmrtznyb1ero8.png)
Now we will substitute -3 for x and 24 for y, we have
![24=a(b)^(-3)](https://img.qammunity.org/2023/formulas/mathematics/college/8vkfuegli6ut3nsaviojeew8jhiebcogv5.png)
Also substituting -2 for x and 12 for y, we have
![12=a(b)^(-2)](https://img.qammunity.org/2023/formulas/mathematics/college/o25qln40aq5svqil2yr45d25y7hcz44btl.png)
Dividing equation 2 by equation 1, we have
![\begin{gathered} (12)/(24)=(a(b)^(-2))/(a(b)^(-3)) \\ (1)/(2)=b^(-2-(-3)) \\ (1)/(2)=b^(-2+3) \\ (1)/(2)=b \\ b=(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ktcrhpc5i3qs88g3hr2rvoxfjgls1jdlfd.png)
Now substitute this value of b into equation 2, we have
![\begin{gathered} 12=a(b)^(-2) \\ 12=a((1)/(2))^(-2) \\ 12=a*(2^(-1))^(-2) \\ 12=a*2^2 \\ 12=4a \\ a=(12)/(4) \\ a=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/j99gjtijg79zt3cjfbed7g1s0kifdegi02.png)
Substituting the values of a for 3 and b for 1/2, we have the equation as
![\begin{gathered} y=a(b)^x \\ y=3((1)/(2))^x \\ y=3(0.5)^x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2fh6g2voh2q1iw9k6em1ecjl55e0ujxsul.png)
Hence the last option is the answer