1.3 It is given that $$ K^{f(x)}=3, K^{g(x)}=4 $$ and $$ K^{\frac{2}{h(x)}}=12 $$ Prove that $$ 2=h(x) \cdot f(x)+h(x) \cdot g(x) $$

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We are given:
$$
K^{f(x)}=3,\quad K^{g(x)}=4,\quad K^{\frac{2}{h(x)}}=12.
$$

**Step 1. Express $$ f(x) $$ in terms of $$\ln$$:**

Taking the natural logarithm on both sides of $$ K^{f(x)}=3 $$, we have
$$
f(x)\ln K = \ln 3 \quad \Longrightarrow \quad f(x)=\frac{\ln 3}{\ln K}.
$$

**Step 2. Express $$ g(x) $$ in terms of $$\ln$$:**

Similarly, for $$ K^{g(x)}=4 $$,
$$
g(x)\ln K = \ln 4 \quad \Longrightarrow \quad g(x)=\frac{\ln 4}{\ln K}.
$$

**Step 3. Express $$ h(x) $$ in terms of $$\ln$$:**

For $$ K^{\frac{2}{h(x)}}=12 $$,
$$
\frac{2}{h(x)}\ln K = \ln 12 \quad \Longrightarrow \quad h(x)=\frac{2\ln K}{\ln 12}.
$$

**Step 4. Compute $$ h(x) \cdot f(x)+h(x) \cdot g(x) $$:**

We factor $$ h(x) $$ as common:
$$
h(x) \cdot f(x)+h(x) \cdot g(x)=h(x) \bigl(f(x)+g(x)\bigr).
$$

Substitute the expressions for $$ f(x) $$, $$ g(x) $$, and $$ h(x) $$:
$$
h(x) \bigl(f(x)+g(x)\bigr)
=\frac{2\ln K}{\ln 12} \left(\frac{\ln 3}{\ln K}+\frac{\ln 4}{\ln K}\right)
=\frac{2\ln K}{\ln 12}\cdot\frac{\ln 3+\ln 4}{\ln K}.
$$

Simplify by canceling $$\ln K$$:
$$
=\frac{2(\ln 3+\ln 4)}{\ln 12}.
$$

**Step 5. Recognize the logarithm property:**

Notice that
$$
\ln 3 + \ln 4 = \ln(3 \times 4) = \ln 12.
$$

Thus, the expression simplifies to:
$$
\frac{2(\ln 12)}{\ln 12} = 2.
$$

**Conclusion:**

We have shown that:
$$
h(x) \cdot f(x)+h(x) \cdot g(x)=2.
$$
$$ h(x) \cdot f(x) + h(x) \cdot g(x) = 2 $$

1.3 It is given that and
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1.3 It is given that and Prove that