You have the following equation:
![((1)/(8))^x=-3](https://img.qammunity.org/2023/formulas/mathematics/college/ajyzrx5fqv0kif6nlb9cmslys9kzf868ku.png)
In order to solve the previous equation for x, proceed as follow:
- write 1/8 as 8⁻¹ and simplify exponents:
![((1)/(8))^x=(8^(-1))^x=8^(-x)](https://img.qammunity.org/2023/formulas/mathematics/college/54pmf2fnn8ic2c6a599quse24kx1rdkqrf.png)
Then, the equation becomes:
![8^(-x)=-3](https://img.qammunity.org/2023/formulas/mathematics/college/5skzpfqdvce77gvos64w70qw53d92gthfb.png)
next, use properties of logarithms to obtain x. In this case apply log with base 8 to cancel 8, as follow:
![\log _8(8^(-x))=-\log _88^x=-x](https://img.qammunity.org/2023/formulas/mathematics/college/reawubypiimvlm6i1rp6nln5edke39fykp.png)
and the equation becomes:
![-x=\log _8(-3)](https://img.qammunity.org/2023/formulas/mathematics/college/5gme02gj0q5mrbn9vmc7aa83l2w9jb4ml2.png)
Then, you obtain log_8 (-3). Due to logarithms of negative numbers do not exist, then, the equation does not have solution for x.
Hence, in both (a) and (b) cases you have:
The solution is the empty set