Surds & Indices Questions and Answers

Answer: C

The product is \(xyz = 1\).

Taking reciprocals, \(p^{-x} = \frac{1}{q}\) gives \(p^{x} = q\); likewise \(q^{y} = r\) and \(r^{z} = p\). Substitute each into the next:

\[ \begin{aligned} r &= q^{y} = (p^{x})^{y} = p^{xy} \\ p &= r^{z} = p^{xyz} \end{aligned} \]

So \(p^{xyz} = p^{1}\), and \(xyz = 1\) for a general base \(p\). The answer \(-1\) comes from forgetting that the negative exponents were cleared by the reciprocals, and \(\frac{1}{pqr}\) and \(pqr\) mix the bases into the exponents, which the equations never do.

Answer: D

(1.001)³³²¹ = 27.64.

The exponents add up: 1259 + 2062 = 3321. Powers of the same base multiply when their exponents add, so:

\[ \begin{aligned} (1.001)^{3321} &= (1.001)^{1259} \times (1.001)^{2062} \\ &= 3.52 \times 7.85 = 27.632 \end{aligned} \]

which agrees with 27.64 within the rounding of the two given values (3.52 and 7.85 are themselves rounded). 11.37 is 3.52 + 7.85, adding the values instead of multiplying them. 4.33 and 2.23 are a difference and a ratio of the two values, which would answer a subtraction of exponents, not an addition.

Answer: D

The value is about 27.64.

The exponents add: \(1259 + 2062 = 3321\). So by the law of indices

\[ \begin{aligned} (1.001)^{3321} &= (1.001)^{1259}\times(1.001)^{2062} \\ &= 3.52 \times 7.85 = 27.632 \end{aligned} \]

which rounds to the option 27.64. 11.37 adds the two values instead of multiplying them, 4.33 is their difference, and 2.23 is their ratio; none follows from the rule \(a^{m}a^{n} = a^{m+n}\).

Answer: C

The value of xyz is 1.

Substitute each relation into the one before it so that everything is expressed as a power of a.

\[ \begin{aligned} a &= b^{x} = (c^{y})^{x} = c^{xy} \\ &= (a^{z})^{xy} = a^{xyz} \end{aligned} \]

Since \(a = a^{1}\), the exponents must match, so xyz = 1 (for a base other than 0 or 1, the usual assumption in such questions).

−1 would give \(a = 1/a\), and 0 would give a = 1 for every a. abc is a product of the bases, not of the exponents, so it is not what the relations determine.

Answer: D

The largest is \(9^{16}\).

Compare the numbers in a common form. \(9^{16} = (3^2)^{16} = 3^{32}\), which is bigger than \(3^{31}\) by a factor of 3.

Take logarithms (base 10) for the other two:

\[ \begin{aligned} \log 3^{32} &= 32 \times 0.4771 \approx 15.27 \\ \log 4^{25} &= 25 \times 0.6021 \approx 15.05 \\ \log 5^{21} &= 21 \times 0.6990 \approx 14.68 \end{aligned} \]

So \(9^{16} \gt 4^{25} \gt 5^{21}\), and \(3^{31}\) is a third of \(9^{16}\). The trap is to assume the largest exponent wins; here the base matters as much as the power.

Answer: B

The value of x is 3.

Simplify each root by taking out square factors:

\[ \begin{aligned} \sqrt{12} + \sqrt{147} &= 2\sqrt{3} + 7\sqrt{3} \\ &= 9\sqrt{3} \end{aligned} \]

So \(9\sqrt{x} = 9\sqrt{3}\) and \(x = 3\). Adding the numbers under the roots (12 + 147 = 159) is the common error and gives none of the options.

Answer: A

\((256)^{0.16}\times (256)^{0.09}=(256)^{(0.16+0.09)}\)

\(=(256)^{0.25}\)

\(=(256)^{(25/100)}\)

\(=(256)^{(1/4)}\)

\(=(4^{4})^{(1/4)}\)

\(=4^{4(1/4)}\)

\(=4^{1}\)

\(=4\)

Answer: C

\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}} \)

\(= \frac{1}{(1+\frac{an}{am})}+\frac{1}{(1+\frac{am}{an})} \)

\(= \frac{am}{(am+an)}+\frac{an}{(am+an)} \)

\(= \frac{(am+an)}{(am+an)} \)

\(= 1\)

Answer: D

\(2^x \times 16^{\frac{2}{5}} = 2^{\frac{1}{5}}\)

\(\Rightarrow 2^x \times (2^4)^{\frac{2}{5}} = 2^{\frac{1}{5}} \)

\(\Rightarrow 2^x \times 2^{\frac{8}{5}}= 2^{\frac{1}{5}} \)

\(\Rightarrow 2^{(x+ \frac{8}{5})} = 2^{\frac{1}{5}} \)

\(\Rightarrow x + \frac{8}{5} = \frac{1}{5} \)

\(\Rightarrow x = (\frac{1}{5} - \frac{8}{5}) = -\frac{7}{5}\)

Answer: B

Let \((25)^{7.5} \times (5)^{2.5} \div (125)^{1.5} = 5x.\)

Then, \(\frac{(5)^{7.5} \times (5)^{2.5}}{(5^{3})^{1.5}} = 5x \)

\(\Rightarrow \frac{5^{(2 \times 7.5)} \times (5)^{2.5}}{5^{(3 \times 1.5)}} = 5x\)

\(\Rightarrow \frac{5^{15} \times 5^{2.5}}{5^{4.5}} = 5x\)

\(\Rightarrow 5x = 5^{(15+2.5-4.5)} \)

\(\Rightarrow 5x = 5^{13}, \therefore x = 13\)