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Number - Logarithms, including laws of logarithms-extended

Grade 10IB

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

🔑Concepts

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A logarithm is the inverse operation of exponentiation. If ay=xa^y = x, then log⁡ax=y\log_a x = y, where a>0,a≠1a > 0, a \neq 1, and x>0x > 0.

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The 'base' of the logarithm is the same as the base of the exponent. The common logarithm uses base 10 (written as log⁡x\log x) and the natural logarithm uses base ee (written as ln⁡x\ln x).

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Logarithms are only defined for positive real numbers. The argument xx in log⁡ax\log_a x must be greater than zero.

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The Change of Base formula allows you to calculate logarithms with any base using a calculator's standard log⁡\log (base 10) or ln⁡\ln (base ee) functions.

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Logarithmic laws are essential for simplifying expressions and solving exponential equations where the bases cannot be made the same.

📐Formulae

log⁡a(xy)=log⁡ax+log⁡ay\log_a (xy) = \log_a x + \log_a y

log⁡a(xy)=log⁡ax−log⁡ay\log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y

log⁡a(xk)=klog⁡ax\log_a (x^k) = k \log_a x

log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a}

log⁡aa=1\log_a a = 1

log⁡a1=0\log_a 1 = 0

alog⁡ax=xa^{\log_a x} = x

💡Examples

Problem 1:

Simplify the expression 2log⁡36−log⁡342 \log_3 6 - \log_3 4 into a single logarithmic term.

Solution:

2log⁡36−log⁡34=log⁡3(62)−log⁡342 \log_3 6 - \log_3 4 = \log_3 (6^2) - \log_3 4 =log⁡336−log⁡34= \log_3 36 - \log_3 4 =log⁡3(364)= \log_3 \left(\frac{36}{4}\right) =log⁡39= \log_3 9 =log⁡3(32)=2= \log_3 (3^2) = 2

Explanation:

First, use the power law to move the coefficient 2 to the exponent of 6. Then, use the quotient law to combine the subtraction into a division. Finally, evaluate the result since 9 is a power of 3.

Problem 2:

Solve for xx: 5x=1205^x = 120. Give your answer to 3 significant figures.

Solution:

log⁡(5x)=log⁡(120)\log(5^x) = \log(120) xlog⁡5=log⁡120x \log 5 = \log 120 x=log⁡120log⁡5x = \frac{\log 120}{\log 5} x≈2.97x \approx 2.97

Explanation:

To solve an exponential equation where the bases cannot be easily equated, take the logarithm of both sides. Use the power law to bring the xx down as a multiplier, then divide to isolate xx.

Problem 3:

Evaluate log⁡832\log_8 32 without using a calculator.

Solution:

Let log⁡832=x\text{Let } \log_8 32 = x 8x=328^x = 32 (23)x=25(2^3)^x = 2^5 23x=252^{3x} = 2^5 3x=53x = 5 x=53x = \frac{5}{3}

Explanation:

Rewrite the logarithmic equation in exponential form. Express both 8 and 32 as powers of the same base (base 2). Equate the exponents and solve for xx.