Exponents

Evaluate each square first: 13^2 = 13 × 13 = 169 and 14^2 = 14 × 14 = 196. Therefore, 13^2 + 14^2 = 169 + 196 = 365, so option B is correct. Exam tip: calculate the powers before performing the addition.

Evaluate the powers first: \(2^7=128\) and \(3^5=243\). Their sum is \(128+243=371\), so option C is correct. An exponent indicates how many times the base is multiplied by itself; calculating each power separately before addition is a reliable exam method.

Compute directly: \(14^2=196\) and \(15^2=225\), so \(196+225=421\). Alternate check: use identity \(a^2+b^2=(a+b)^2-2ab\); here \((14+15)^2-2\cdot14\cdot15=29^2-420=841-420=421\). The nearest wrong choice 411 can result from mistakenly taking \(15^2=215\) (giving 196+215=411). Exam tip: memorize common squares and use the identity to reduce calculation errors on paper.

Using the power-of-a-power rule, (a^m)^n = a^(mn), we get (2^4)^2 = 2^(4×2) = 2^8 = 256. Therefore, 256 is correct. The values 128 and 64 correspond to different powers, while 512 is 2^9. Exam tip: when a power is raised to another power, multiply the exponents.

First calculate the squares: 15^2 = 225 and 16^2 = 256. Adding them gives 225 + 256 = 481, so 481 is correct. Remember to evaluate each exponent before performing the addition.

Since (31)^2 = 31 × 31 = 961, the positive value of c is 31. Although c^2 = 961 gives c = 31 or c = -31, the question specifically asks for the positive value. Exam tip: when a positive value is requested from a square, choose the positive square root.

When powers with the same base are divided, their exponents are subtracted: \(a^m \div a^n = a^{m-n}\). Thus, \(3^6 \div 3^3 = 3^{6-3} = 3^3 = 27\). Option B results from incorrectly using \(3^2\). In an exam, remember to subtract exponents when dividing powers with the same base.

First calculate both squares: \(17^2=289\) and \(18^2=324\). Therefore, \(17^2+18^2=289+324=613\), so option B is correct. Exam tip: Calculate each square separately before adding to avoid arithmetic errors.

Evaluate the two powers separately: 4^5 = 1024 and 2^8 = 256. Therefore, 4^5 + 2^8 = 1024 + 256 = 1280, so option C is correct. In an exam, calculate each exponent before performing the addition to avoid confusing the bases or exponents.

Given d^3 = 5832. Since 18 × 18 × 18 = 5832, the value of d is 18, so option A is correct. For an equation involving a cube, find the unknown by taking the cube root: d = ∛5832.

First calculate the two squares: 19^2 = 361 and 20^2 = 400. Therefore, 19^2 + 20^2 = 361 + 400 = 761, so option B is correct. In an exam, calculate each exponent separately before adding the terms to avoid confusing addition of squares with the square of a sum.

For the same base, exponents are subtracted when dividing: \(a^m \div a^n = a^{m-n}\). Thus \(5^5 \div 5^2 = 5^{5-2} = 5^3 = 125\). Closest distractors: 25 is \(5^2\) (too small), 625 is \(5^4\) (result of a wrong operation like adding exponents), and 5 is \(5^1\) (over-subtracted). Exam tip: when dividing like bases, subtract the exponents — don’t divide the bases themselves.

Compute directly: \(20^2 = 400\). Use the identity for the nearby square: \(21^2 = (20+1)^2 = 20^2 + 2\times20\times1 + 1^2 = 400 + 40 + 1 = 441\). Adding gives \(400 + 441 = 841\), so the correct answer is 841.
The closest wrong choice (831) typically arises from a slip in calculating \(21^2\) or an arithmetic error when adding. Exam tip: use \((a+b)^2\) or the nearest perfect square method to avoid careless mistakes.

Using the power-of-a-power rule, \((a^m)^n=a^{mn}\). Therefore, \((4^2)^3=4^{2\times3}=4^6=4096\), so option A is correct. In the exam, remember to multiply the exponents; option B can result from incorrectly taking the exponent as 5 instead of 6.

First calculate the squares: 21^2 = 441 and 22^2 = 484. Therefore, 21^2 + 22^2 = 441 + 484 = 925, so option B is correct. In an exam, calculating each power separately before adding helps avoid arithmetic errors.

Compute each power first: \(6^4=6\times6\times6\times6=1296\) and \(3^5=3\times3\times3\times3\times3=243\). Adding gives \(1296+243=1539\). Option B (1536) is a close arithmetic mistake (treating 243 as 240). Option A (1296) is just \(6^4\) alone. Exam tip: evaluate powers separately and then add; checking the units digit often catches small addition errors.

Direct calculation: \(22^2=484\) and \(23^2=529\). Adding gives \(484+529=1013\).
Alternate quick method (identity): \(a^2+b^2=(a+b)^2-2ab\). So \((22+23)^2-2\times22\times23=45^2-1012=2025-1012=1013\.
Exam tip: Use this identity for fast mental calculation of sums of squares. Option B (1003) is a common slip (off by 10) and therefore incorrect.

When powers with the same base are divided, their exponents are subtracted: \(a^m \div a^n = a^{m-n}\). Therefore, \(7^4 \div 7^1 = 7^{4-1} = 7^3 = 343\). Option B is the value of only \(7^2\), so it is incorrect. Exam tip: In division of powers with the same base, keep the base unchanged and subtract the exponents.

Since (34)^2 = 34 × 34 = 1156, e = ±34. However, the question asks for the positive value, so e = 34. Remember that the positive square root is selected when a positive value is requested; therefore, −34 is not the answer.

Direct method: \(23^2=529\) and \(24^2=576\). Adding gives \(529+576=1105\).
Alternate quick method: use \(a^2+b^2=(a+b)^2-2ab\). Here \((23+24)^2=47^2=2209\) and \(2\times23\times24=1104\), so \(2209-1104=1105\).
Why other options are wrong: 1100 and 1095 result from small arithmetic undercounts; 1115 is an overestimate.
Exam tip: Use the identity \((a+b)^2-2ab\) to avoid computing two large squares separately — it's faster and reduces errors.