Comprehensive study of the applications of logarithms in CBSE Class 11 Applied Mathematics, covering compound interest, depreciation, growth rates, and solving exponential equations with step-by-step examples.
Logarithms are primarily used to solve for time (n) or rate (r) in problems involving compound interest or depreciation, where the variable is present in the exponent. Given the formula A = P(1 + r/100)^n, applying logarithms allows us to convert the exponential relationship into a linear one, making calculation simpler using log tables or calculators.
• Identify the formula: A = P(1 + r)^n. Here, A=7500, P=5000, r=0.10.
• Substitute values: 7500 = 5000(1.1)^n.
• Divide by 5000: 1.5 = (1.1)^n.
• Take natural logarithm (ln) on both sides: ln(1.5) = n * ln(1.1).
• Solve for n: n = ln(1.5) / ln(1.1) approx 0.4055 / 0.0953.
• Calculate the final value: n = 4.25 years.
Answer: The amount will grow in approximately 4.25 years.
When solving equations where the variable is in the exponent, we utilize the property that log(a^b) = b * log(a). This property effectively 'brings down' the exponent, transforming complex exponential equations into simpler algebraic equations that can be solved using standard linear or quadratic methods.
• Take log on both sides: log(3^(x+1)) = log(5^x).
• Apply the power rule: (x + 1) * log(3) = x * log(5).
• Expand the expression: x*log(3) + log(3) = x*log(5).
• Isolate x: x(log(5) - log(3)) = log(3).
• Final form: x = log(3) / (log(5) - log(3)).
Answer: x = log(3) / log(5/3) which is approximately 2.15.