1995 (2) Solve for x in 3log10 x + log10 3 = log10 81

Let's solve these problems step by step.

Problem (2): Solve for $x$ in $3\log_{10} x + \log_{10} 3 = \log_{10} 81$

Step 1: Apply the power rule of logarithms, $a \log b = \log b^a$, to the first term. $\log_{10} x^3 + \log_{10} 3 = \log_{10} 81$

Step 2: Apply the product rule of logarithms, $\log a + \log b = \log (ab)$, to the left side of the equation. $\log_{10} (x^3 \times 3) = \log_{10} 81$ $\log_{10} (3x^3) = \log_{10} 81$

Step 3: Since the logarithms have the same base and are equal, their arguments must be equal. $3x^3 = 81$

Step 4: Solve for $x$. $x^3 = \frac{81}{3}$ $x^3 = 27$ Take the cube root of both sides: $x = \sqrt[3]{27}$ $x = 3$ The value of $x$ is $\boxed{3}$.

Problem (3) (a): If $3^m \times 3^n = 243$ and $3^m \div 3^{2n} = 9$, find $m$ and $n$.

Step 1: Simplify the first equation using the exponent rule $a^b \times a^c = a^{b+c}$. $3^{m+n} = 243$ Express $243$ as a power of $3$: $3^5 = 243$. $3^{m+n} = 3^5$ Equating the exponents, we get our first linear equation: $m+n = 5 \quad \text{(Equation 1)}$

Step 2: Simplify the second equation using the exponent rule $a^b \div a^c = a^{b-c}$. $3^{m-2n} = 9$ Express $9$ as a power of $3$: $3^2 = 9$. $3^{m-2n} = 3^2$ Equating the exponents, we get our second linear equation: $m-2n = 2 \quad \text{(Equation 2)}$

Step 3: Solve the system of linear equations (Equation 1 and Equation 2). Subtract Equation 2 from Equation 1: $(m+n) - (m-2n) = 5 - 2$ $m+n-m+2n = 3$ $3n = 3$ $n = 1$

Step 4: Substitute the value of $n$ into Equation 1 to find $m$. $m+1 = 5$ $m = 5-1$ $m = 4$ The values are $m = \boxed{4}$ and $n = \boxed{1}$.