Exponents and logarithms: inverse operations Test

{"questions":[{"content":"Enter $5^6/5^4$. [[input-exponent-logarithm-laws-test-1]]","widgets":{"input-exponent-logarithm-laws-test-1":{"type":"input","answer":"25"}},"hints":[],"explanation":"Subtract the exponents to obtain 5².","id":"exponent-logarithm-laws-test-1","coverage_topics":["aa-hl-laws-of-exponents"],"correct_feedback":"Subtract the exponents to obtain 5².","incorrect_feedback":"Subtract the exponents to obtain 5².","evidence_kind":"objective-concept-test"},{"content":"For $a\\gt0$, simplify $\\sqrt{a^6}/a^{-2}$.\u003Cbr\u003E[[choice-exponent-logarithm-laws-test-2]]","widgets":{"choice-exponent-logarithm-laws-test-2":{"type":"choice","options":["$a$","$a^8$","$a^5$"],"answer":[2],"choice_feedback":["Dividing by a⁻² subtracts −2, not 2.","Taking the square root halves the exponent 6 before the division.","The positive-base condition gives √(a⁶) = a³, and dividing by a⁻² adds 2 to the exponent."]}},"hints":[],"explanation":"The positive-base condition gives √(a⁶) = a³, and dividing by a⁻² adds 2 to the exponent.","id":"exponent-logarithm-laws-test-2","coverage_topics":["aa-hl-laws-of-exponents"],"correct_feedback":"The positive-base condition gives √(a⁶) = a³, and dividing by a⁻² adds 2 to the exponent.","incorrect_feedback":"The positive-base condition gives √(a⁶) = a³, and dividing by a⁻² adds 2 to the exponent.","evidence_kind":"objective-concept-test"},{"content":"For $a\\gt 0$, simplify $(a^{-2})^3a^5$.\u003Cbr\u003E[[choice-exponent-logarithm-laws-test-3]]","widgets":{"choice-exponent-logarithm-laws-test-3":{"type":"choice","options":["$a^{-30}$","$a^{-1}$","$a^6$"],"answer":[1],"choice_feedback":["Multiplying exponents applies to nested powers, not the outside product.","Multiply exponents inside the power, then add: −6 + 5 = −1.","Adding −2 and 3 is not the power-of-a-power rule."]}},"hints":[],"explanation":"Multiply exponents inside the power, then add: −6 + 5 = −1.","id":"exponent-logarithm-laws-test-3","coverage_topics":["aa-hl-laws-of-exponents"],"correct_feedback":"Multiply exponents inside the power, then add: −6 + 5 = −1.","incorrect_feedback":"Multiply exponents inside the power, then add: −6 + 5 = −1.","evidence_kind":"objective-concept-test"},{"content":"Enter $\\log_2(1/32)$. [[input-exponent-logarithm-laws-test-4]]","widgets":{"input-exponent-logarithm-laws-test-4":{"type":"input","answer":"-5"}},"hints":[],"explanation":"A logarithm returns the exponent: 2⁻⁵ = 1/32, so the answer is −5.","id":"exponent-logarithm-laws-test-4","coverage_topics":["aa-hl-laws-of-logarithms"],"correct_feedback":"A logarithm returns the exponent: 2⁻⁵ = 1/32, so the answer is −5.","incorrect_feedback":"A logarithm returns the exponent: 2⁻⁵ = 1/32, so the answer is −5.","evidence_kind":"objective-concept-test"},{"content":"Simplify $2\\ln x-\\ln(x-3)$ on its real domain.\u003Cbr\u003E[[choice-exponent-logarithm-laws-test-5]]","widgets":{"choice-exponent-logarithm-laws-test-5":{"type":"choice","options":["$\\ln(2x/(x-3)),\\ x\\gt 3$","$\\ln(x^2-x+3),\\ x\\gt 3$","$\\ln\\!\\left(x^2/(x-3)\\right),\\ x\\gt 3$"],"answer":[2],"choice_feedback":["The coefficient becomes an exponent on x, not a factor 2x.","Subtracting logs divides the arguments.","Use the power and quotient rules while retaining both original positive-argument conditions."]}},"hints":[],"explanation":"Use the power and quotient rules while retaining both original positive-argument conditions.","id":"exponent-logarithm-laws-test-5","coverage_topics":["aa-hl-laws-of-logarithms"],"correct_feedback":"Use the power and quotient rules while retaining both original positive-argument conditions.","incorrect_feedback":"Use the power and quotient rules while retaining both original positive-argument conditions.","evidence_kind":"objective-concept-test"},{"content":"If $\\log_3 a=2$ and $\\log_3 b=4$, enter $\\log_3(a^2/b)$. [[input-exponent-logarithm-laws-test-6]]","widgets":{"input-exponent-logarithm-laws-test-6":{"type":"input","answer":"0"}},"hints":[],"explanation":"The logarithm is 2(2) − 4 = 0.","id":"exponent-logarithm-laws-test-6","coverage_topics":["aa-hl-laws-of-logarithms"],"correct_feedback":"The logarithm is 2(2) − 4 = 0.","incorrect_feedback":"The logarithm is 2(2) − 4 = 0.","evidence_kind":"objective-concept-test"},{"content":"Which expression equals $\\log_{1/2}3$?\u003Cbr\u003E[[choice-exponent-logarithm-laws-test-7]]","widgets":{"choice-exponent-logarithm-laws-test-7":{"type":"choice","options":["$-\\ln3/\\ln2$","$\\ln3/\\ln2$","$\\ln2/\\ln3$"],"answer":[0],"choice_feedback":["ln(1/2) = −ln 2, so the change-of-base ratio is negative.","A base below 1 gives a negative exponent for an argument above 1.","This reverses the requested base and argument."]}},"hints":[],"explanation":"ln(1/2) = −ln 2, so the change-of-base ratio is negative.","id":"exponent-logarithm-laws-test-7","coverage_topics":["aa-hl-laws-of-logarithms"],"correct_feedback":"ln(1/2) = −ln 2, so the change-of-base ratio is negative.","incorrect_feedback":"ln(1/2) = −ln 2, so the change-of-base ratio is negative.","evidence_kind":"objective-concept-test"},{"content":"A scale is defined by $L=\\log_{10}(I/I_0)$, with fixed positive $I_0$. Two readings differ by 2. Enter the ratio of the larger intensity to the smaller intensity. [[input-exponent-logarithm-laws-test-8]]","widgets":{"input-exponent-logarithm-laws-test-8":{"type":"input","answer":"100"}},"hints":[],"explanation":"Subtract the two readings: log base 10 of the intensity ratio is 2. The ratio is therefore 10².","id":"exponent-logarithm-laws-test-8","coverage_topics":["aa-hl-laws-of-logarithms"],"correct_feedback":"Subtract the two readings: log base 10 of the intensity ratio is 2. The ratio is therefore 10².","incorrect_feedback":"Subtract the readings and combine the logarithms before converting back to an intensity ratio.","evidence_kind":"objective-concept-test"}],"mix":0,"tinkumo_module_mode":"test"}