Solve for $$ x $$ $$ \begin{array}{ll} ext { 1.1 } \frac{1}{\frac{5}{3}}=32 & ext { (4) } \ x^{\frac{3}{3}} \ ext { \&2. } \sqrt{x-5}-x=3 & ext { (6) } \ 2.3 \log _{x}=-1 & ext { (2) }\end{array} $$

Question

Solve for $$ x $$ $$ \begin{array}{ll}	ext { 1.1 } \frac{1}{\frac{5}{3}}=32 & 	ext { (4) } \ x^{\frac{3}{3}} \ 	ext { \&2. } \sqrt{x-5}-x=3 & 	ext { (6) } \ 2.3 \log _{x}=-1 & 	ext { (2) }\end{array} $$

UpStudy AI · Accepted Answer

Let's solve the equations one by one.

### Equation 1: 
$$
\frac{1}{\frac{5}{3}} = 32
$$
This equation simplifies to:
$$
\frac{3}{5} = 32
$$
This is not true, so there is no solution from this equation.

### Equation 2:
$$
\sqrt{x-5} - x = 3
$$
Rearranging gives:
$$
\sqrt{x-5} = x + 3
$$
Now, squaring both sides:
$$
x - 5 = (x + 3)^2
$$
Expanding the right side:
$$
x - 5 = x^2 + 6x + 9
$$
Rearranging gives:
$$
0 = x^2 + 5x + 14
$$
Now we can use the quadratic formula to solve for $$ x $$:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1}
$$
Calculating the discriminant:
$$
5^2 - 4 \cdot 1 \cdot 14 = 25 - 56 = -31
$$
Since the discriminant is negative, there are no real solutions for this equation.

### Equation 3:
$$
2.3 \log_x = -1
$$
This equation seems to be missing a term after $$ \log_x $$. Assuming it should be $$ \log_x(y) $$ for some $$ y $$, we cannot solve it without additional information.

### Conclusion:
From the equations provided, we found that:
1. The first equation has no solution.
2. The second equation has no real solutions.
3. The third equation is incomplete.

Thus, there are no solutions for $$ x $$ based on the provided equations. If you have more context or corrections, please provide them for further assistance!
The equations provided do not have solutions for $$ x $$.

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