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Logarithms - Logarithms Across Subjects-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition: A logarithm is the inverse operation to exponentiation. If by=xb^y = x, then log⁡b(x)=y\log_b(x) = y, where b>0b > 0 and b≠1b \neq 1.

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Common Logarithms: Logarithms with base 1010, written as log⁡10x\log_{10} x or simply log⁡x\log x, are used extensively in scientific scales.

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Natural Logarithms: Logarithms with base ee (Euler's number ≈2.718\approx 2.718), written as ln⁡x\ln x, used in growth and decay models.

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Chemistry (pH Scale): The acidity of a solution is measured by the concentration of hydrogen ions [H+][H^+]. The pH is defined as the negative base-1010 logarithm of this concentration.

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Physics (Sound Intensity): Sound levels are measured in Decibels (dBdB), which is a logarithmic unit comparing a sound's intensity to a reference intensity.

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Geology (Richter Scale): The magnitude of an earthquake is determined using a base-1010 logarithmic scale to represent the vast range of energy released.

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Finance: Logarithms are used to solve for the time period nn in the compound interest formula A=P(1+r)nA = P(1 + r)^n when AA, PP, and rr are known.

📐Formulae

log⁡b(x)=y  ⟺  by=x\log_b(x) = y \iff b^y = x

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

LdB=10log⁡10(II0)L_{dB} = 10 \log_{10}\left(\frac{I}{I_0}\right) where I0=10−12W/m2I_0 = 10^{-12} W/m^2

M=log⁡10(AA0)M = \log_{10}\left(\frac{A}{A_0}\right) (Richter Magnitude)

n=log⁡(A/P)log⁡(1+r)n = \frac{\log(A/P)}{\log(1 + r)} (Time in Compound Interest)

💡Examples

Problem 1:

Calculate the pH of a lemon juice solution if the hydrogen ion concentration [H+][H^+] is 0.0010.001 moles per liter.

Solution:

Given [H+]=0.001=10−3[H^+] = 0.001 = 10^{-3}. Using the formula pH=−log⁡10[H+]pH = -\log_{10}[H^+]: pH=−log⁡10(10−3)pH = -\log_{10}(10^{-3}) pH=−(−3)log⁡10(10)pH = -(-3) \log_{10}(10) Since log⁡10(10)=1\log_{10}(10) = 1: pH=3pH = 3

Explanation:

We express the concentration as a power of 1010 and apply the power rule of logarithms log⁡(an)=nlog⁡a\log(a^n) = n \log a.

Problem 2:

A sound has an intensity II of 10−7W/m210^{-7} W/m^2. Find the sound level in decibels (dBdB) given the reference intensity I0=10−12W/m2I_0 = 10^{-12} W/m^2.

Solution:

Use the formula L=10log⁡10(II0)L = 10 \log_{10}\left(\frac{I}{I_0}\right). L=10log⁡10(10−710−12)L = 10 \log_{10}\left(\frac{10^{-7}}{10^{-12}}\right) L=10log⁡10(10−7−(−12))L = 10 \log_{10}(10^{-7 - (-12)}) L=10log⁡10(105)L = 10 \log_{10}(10^5) L=10×5×log⁡10(10)L = 10 \times 5 \times \log_{10}(10) L=50dBL = 50 dB

Explanation:

The ratio of intensities is calculated first using laws of exponents, then the base-1010 logarithm is applied.

Problem 3:

In how many years will Rs 1000 double itself at an annual interest rate of 10%10\% compounded annually? (Given log⁡2≈0.3010\log 2 \approx 0.3010 and log⁡1.1≈0.0414\log 1.1 \approx 0.0414)

Solution:

We use the compound interest formula A=P(1+r)nA = P(1 + r)^n. For the amount to double, A=2PA = 2P. 2P=P(1+0.10)n2P = P(1 + 0.10)^n 2=(1.1)n2 = (1.1)^n Taking log⁡\log on both sides: log⁡(2)=log⁡(1.1n)\log(2) = \log(1.1^n) log⁡(2)=nlog⁡(1.1)\log(2) = n \log(1.1) n=log⁡2log⁡1.1n = \frac{\log 2}{\log 1.1} n=0.30100.0414≈7.27n = \frac{0.3010}{0.0414} \approx 7.27 years.

Explanation:

To solve for an exponent (nn), we apply logarithms to both sides of the equation and use the power rule.