Indices
Indices (also known as exponents or powers) represent how many times a number or variable (the base) is multiplied by itself. For example, in \(2^3\), 2 is the base, and 3 is the index (or exponent), meaning \(2 \times 2 \times 2 = 8\).
\(\text{So, } a \times a \times a \times a = a^4\)
For Basic Formula on Indices Click Here
Exponential Equation
An exponential equation is an equation in which the variable appears in the exponent (the power). Example: \(3^{(x+2)} = 81\).
Some formulae for solving exponential equation.
1. If the bases are the same, then the powers will be equal.
i.e \((a^x = a^y)\) then, \((x = y)\). \((a \neq 0,1)\)
2. If \((a^x = b^y)\) then, \((a = b^{\frac{y}{x}})\) and \((b = a^{\frac{x}{y}})\).
Worked out Questions
\(
\begin{aligned}
& \text{Q.1. Solve: } 3^{(x-1)} = 81 \\
& \text{Solution,} \\
& \text{or, } 3^{(x-1)} = 81 \\
& \text{or, } 3^{(x-1)} = 3^4 \\
& \text{or, } x-1 = 4 \\
& \therefore x = 5 \text{ (Answer)}
\end{aligned}
\)
\(\begin{aligned}
& \text{Q. 2. Solve: } 3^{(x+1)} + 3^x = \frac{4}{81} \\
& \text{Solution,} \\
& \text{or, } 3^{(x+1)} + 3^x = \frac{4}{81} \\
& \text{or, } 3^x + 3^1 + 3^x = \frac{4}{81} \\
& \text{or, } 3^x (3 + 1) = \frac{4}{81} \\
& \text{or, } 3^x (4) = \frac{4}{81} \\
& \text{or, } 3^x = \frac{1}{81} \\
& \text{or, } 3^x = 3^{-4} \\
& \therefore x = -4 \text{ (Answer)}\\
\end{aligned}\)
\(\begin{aligned}
& \text{Q. 3. Solve: } 2^x + \frac{1}{2^x} = 2\frac{1}{2} \\
& \text{Solution,} \\
& \text{or, } 2^x + \frac{1}{2^x} = 2\frac{1}{2} \\
& \text{or, } \text{Let us consider, } 2^x = a \\
& \text{or, } a + \frac{1}{a} = \frac{5}{2} \\
& \text{or, } \frac{a^2 +1}{a} = \frac{5}{2} \\
& \text{or, } 2a^2 + 2 = 5a \\
& \text{or, } 2a^2 - 5a + 2 = 0 \\
& \text{or, } 2a^2 - (4 + 1)a + 2 = 0 \\
& \text{or, } 2a^2 - 4a - 1a + 2 = 0 \\
& \text{or, } 2a(a - 2) - 1(a - 2) = 0 \\
& \text{or, } (a - 2)(2a - 1) = 0 \\
& \text{Either, } a - 2 = 0 \text{ or, } 2a - 1 = 0 \\
& \text{or, } a = 2 \text{ or, } a = \frac{1}{2} \\
& \text{or, } 2^x = 2 \text{ or, } 2^x = \frac{1}{2} \\
& \text{or, } 2^x = 2^1 \text{ or, } 2^x = 2^{-1} \\
& \therefore x = 1 \text{ or, } x = -1 \\
& \text{Hence, } x = \pm 1 \text{ (Answer)}
\end{aligned}\)
\(\begin{aligned}
& \text{Q. 4. Solve: } 4 \times 3^{(x + 1)} – 9^x = 27 \\
& \text{Solution,} \\
& \text{or, } 4 \times 3^{(x + 1)} – 9^x = 27 \\
& \text{or, } 4 \times 3^x \times 3^1 – 3^2x = 27 \\
& \text{Now, } \text{Let us consider, } 3^x = a \\
& \text{or, } 12a – a^2 = 27 \\
& \text{or, } a^2 - 12a + 27 = 0 \\
& \text{or, } a^2 – (9 + 3)a + 27 = 0 \\
& \text{or, } a^2 – 9a - 3a + 27 = 0 \\
& \text{or, } a (a – 9) – 3(a – 9) = 0 \\
& \text{or, } (a – 9) (a – 3) = 0 \\
& \text{Either, } (a – 9) = 0 \quad \text{ or, } (a – 3) = 0 \\
&\text{or, } a = 9 \quad \text{ or, } \quad a = 3 \\
& \text{or, } 3^x = 3^2 \quad \text{ or, } \quad 3^x = 3^1 \\
& \therefore x = 2 \quad \text{ or, } \quad x = 1 \\
& \text{Hence, } x = 2 \quad \text{ or } \quad 1 \text{ (Answer)}
\end{aligned}\)
Class 10: Mathematics
