Solve the following equations
`(i) (log_(2)(9-2^(x)))/(3-x)=1`
`(ii) x^((log_(10)x+7)/(4))=10^((log_(10)x+1)`

Answer

Let's solve the given equations step by step.

### Part (i): Solve the equation
\[ \frac{\log_{2}(9 - 2^{x})}{3 - x} = 1 \]

**Step 1: Isolate the logarithm.**
Multiply both sides by \(3 - x\):
\[ \log_{2}(9 - 2^{x}) = 3 - x \]

**Step 2: Convert the logarithmic equation to its exponential form.**
Using the property of logarithms, if \(\log_{b}(a) = c\), then \(a = b^{c}\):
\[ 9 - 2^{x} = 2^{3 - x} \]

**Step 3: Rearrange the equation.**
Rearranging gives:
\[ 9 - 2^{x} = 2^{3} \cdot 2^{-x} = 8 \cdot 2^{-x} \]
This can be rewritten as:
\[ 9 - 2^{x} = \frac{8}{2^{x}} \]

**Step 4: Multiply through by \(2^{x}\) to eliminate the fraction.**
\[ (9 - 2^{x}) \cdot 2^{x} = 8 \]
Expanding gives:
\[ 9 \cdot 2^{x} - (2^{x})^{2} = 8 \]
This simplifies to:
\[ 9 \cdot 2^{x} - 2^{2x} - 8 = 0 \]

**Step 5: Let \(y = 2^{x}\).**
Substituting gives:
\[ 9y - y^{2} - 8 = 0 \]
Rearranging gives:
\[ y^{2} - 9y + 8 = 0 \]

**Step 6: Factor the quadratic equation.**
Factoring gives:
\[ (y - 1)(y - 8) = 0 \]
Thus, \(y = 1\) or \(y = 8\).

**Step 7: Solve for \(x\).**
Recall \(y = 2^{x}\):
1. If \(y = 1\), then \(2^{x} = 1 \Rightarrow x = 0\).
2. If \(y = 8\), then \(2^{x} = 8 \Rightarrow x = 3\).

**Final solutions for part (i):**
\[ x = 0 \quad \text{or} \quad x = 3 \]

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### Part (ii): Solve the equation
\[ x^{\frac{\log_{10}x + 7}{4}} = 10^{\log_{10}x + 1} \]

**Step 1: Rewrite the right side.**
The right side can be rewritten as:
\[ 10^{\log_{10}x + 1} = 10^{\log_{10}x} \cdot 10^{1} = x \cdot 10 \]
Thus, the equation becomes:
\[ x^{\frac{\log_{10}x + 7}{4}} = 10x \]

**Step 2: Divide both sides by \(x\) (assuming \(x \neq 0\)).**
\[ x^{\frac{\log_{10}x + 7}{4} - 1} = 10 \]

**Step 3: Simplify the exponent.**
The exponent simplifies to:
\[ \frac{\log_{10}x + 7 - 4}{4} = \frac{\log_{10}x + 3}{4} \]
Thus, we have:
\[ x^{\frac{\log_{10}x + 3}{4}} = 10 \]

**Step 4: Rewrite in logarithmic form.**
Taking logarithm base 10 of both sides:
\[ \frac{\log_{10}x + 3}{4} \cdot \log_{10}x = 1 \]

**Step 5: Multiply both sides by 4.**
\[ \log_{10}x \cdot (\log_{10}x + 3) = 4 \]

**Step 6: Let \(t = \log_{10}x\).**
Then we have:
\[ t(t + 3) = 4 \]
This expands to:
\[ t^{2} + 3t - 4 = 0 \]

**Step 7: Factor the quadratic equation.**
Factoring gives:
\[ (t - 1)(t + 4) = 0 \]
Thus, \(t = 1\) or \(t = -4\).

**Step 8: Solve for \(x\).**
Recall \(t = \log_{10}x\):
1. If \(t = 1\), then \(x = 10^{1} = 10\).
2. If \(t = -4\), then \(x = 10^{-4} = 0.0001\).

**Final solutions for part (ii):**
\[ x = 10 \quad \text{or} \quad x = 0.0001 \]

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