Indices and Logarithms Quiz - ICSE Class 10 Mathematics
Free practice quiz: Indices and Logarithms (ICSE Class 10 Mathematics). 47 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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If $a^x = b$, $b^y = c$, and $c^z = a$, then find the value of $xyz$.
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Answer: A. 1
Given $a^x = b$. Substituting this into $b^y = c$, we get $(a^x)^y = c$, which simplifies to $a^{xy} = c$. Substituting this into $c^z = a$, we get $(a^{xy})^z = a$, which simplifies to $a^{xyz} = a^1$. Therefore, $xyz = 1$.
If $9^{x+1} = 27^{x-2}$, find $x$.
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Answer: B. 8
We can rewrite both 9 and 27 as powers of 3. $9 = 3^2$ and $27 = 3^3$. So the equation becomes $(3^2)^{x+1} = (3^3)^{x-2}$. This simplifies to $3^{2(x+1)} = 3^{3(x-2)}$. Equating the exponents: $2(x+1) = 3(x-2)$. $2x+2 = 3x-6$. Subtracting $2x$ from both sides gives $2 = x-6$. Adding 6 to both sides gives $x = 8$.
If $\log_a 25 = 2$, what is $a$?
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Answer: C. 5
The equation $\log_a 25 = 2$ means $a^2 = 25$. Taking the square root of both sides, we get $a = \pm 5$. However, the base of a logarithm must be positive and not equal to 1. Therefore, we choose the positive value, $a=5$.
Evaluate: $\log_3 27 - \log_3 3$.
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Answer: B. 2
We know that $27 = 3^3$, so $\log_3 27 = 3$. Also, $\log_3 3 = 1$. Therefore, $\log_3 27 - \log_3 3 = 3 - 1 = 2$. Alternatively, using the property $\log_b m - \log_b n = \log_b (m/n)$, we have $\log_3 27 - \log_3 3 = \log_3 (27/3) = \log_3 9$. Since $9 = 3^2$, $\log_3 9 = 2$.
If $100^x = 1000$, find $x$.
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Answer: B. 3/2
We can rewrite both 100 and 1000 as powers of 10. $100 = 10^2$ and $1000 = 10^3$. So the equation becomes $(10^2)^x = 10^3$. This simplifies to $10^{2x} = 10^3$. Equating the exponents: $2x = 3$. Dividing by 2: $x = 3/2$.