Solving exponential and logarithmic equations Test

{"questions":[{"content":"Solve $25^{x-1}=125$. [[input-exponential-logarithmic-equations-test-1]]","widgets":{"input-exponential-logarithmic-equations-test-1":{"type":"input","answer":"2.5"}},"hints":[],"explanation":"Write both sides with base 5: 2x − 2 = 3.","id":"exponential-logarithmic-equations-test-1","coverage_topics":["aa-hl-exponential-equations"],"correct_feedback":"Write both sides with base 5: 2x − 2 = 3.","incorrect_feedback":"Write both sides with base 5: 2x − 2 = 3.","evidence_kind":"objective-concept-test"},{"content":"Solve $3e^{2x}=15$ exactly.\u003Cbr\u003E[[choice-exponential-logarithmic-equations-test-2]]","widgets":{"choice-exponential-logarithmic-equations-test-2":{"type":"choice","options":["$\\ln5$","$\\ln(15)/6$","$\\tfrac12\\ln5$"],"answer":[2],"choice_feedback":["The exponent is 2x, so divide by 2.","First isolate the exponential by dividing 15 by 3.","Divide by 3, then take ln: 2x = ln5."]}},"hints":[],"explanation":"Divide by 3, then take ln: 2x = ln5.","id":"exponential-logarithmic-equations-test-2","coverage_topics":["aa-hl-exponential-equations"],"correct_feedback":"Divide by 3, then take ln: 2x = ln5.","incorrect_feedback":"Divide by 3, then take ln: 2x = ln5.","evidence_kind":"objective-concept-test"},{"content":"Solve $\\log_5(2x+1)=2$. [[input-exponential-logarithmic-equations-test-3]]","widgets":{"input-exponential-logarithmic-equations-test-3":{"type":"input","answer":"12"}},"hints":[],"explanation":"2x + 1 = 25, hence x = 12.","id":"exponential-logarithmic-equations-test-3","coverage_topics":["aa-hl-logarithmic-equations"],"correct_feedback":"2x + 1 = 25, hence x = 12.","incorrect_feedback":"2x + 1 = 25, hence x = 12.","evidence_kind":"objective-concept-test"},{"content":"Solve $\\ln(x+1)-\\ln(x-1)=\\ln3$. [[input-exponential-logarithmic-equations-test-4]]","widgets":{"input-exponential-logarithmic-equations-test-4":{"type":"input","answer":"2"}},"hints":[],"explanation":"For x \u0026gt; 1, the quotient rule gives (x + 1)/(x − 1) = 3. Thus x = 2, which satisfies the domain.","id":"exponential-logarithmic-equations-test-4","coverage_topics":["aa-hl-logarithmic-equations"],"correct_feedback":"For x \u0026gt; 1, the quotient rule gives (x + 1)/(x − 1) = 3. Thus x = 2, which satisfies the domain.","incorrect_feedback":"First require both arguments to be positive. A difference of logarithms becomes the logarithm of a quotient.","evidence_kind":"objective-concept-test"},{"content":"For which real c does $\\ln(x-1)=c$ have a real solution?\u003Cbr\u003E[[choice-exponential-logarithmic-equations-test-5]]","widgets":{"choice-exponential-logarithmic-equations-test-5":{"type":"choice","options":["Only $c\\gt0$","Only $c\\ge1$","Every real c"],"answer":[2],"choice_feedback":["A logarithm may be negative; its argument, not its output, must be positive.","The restriction x \u0026gt; 1 concerns x, not the logarithm value c.","Rewriting gives x = 1 + eᶜ. Since eᶜ is positive for every real c, the original argument is always valid."]}},"hints":[],"explanation":"Rewriting gives x = 1 + eᶜ. Since eᶜ is positive for every real c, the original argument is always valid.","id":"exponential-logarithmic-equations-test-5","coverage_topics":["aa-hl-logarithmic-equations"],"correct_feedback":"Rewriting gives x = 1 + eᶜ. Since eᶜ is positive for every real c, the original argument is always valid.","incorrect_feedback":"Rewriting gives x = 1 + eᶜ. Since eᶜ is positive for every real c, the original argument is always valid.","evidence_kind":"objective-concept-test"},{"content":"Solve $4^x+2^x-6=0$.\u003Cbr\u003E[[choice-exponential-logarithmic-equations-test-6]]","widgets":{"choice-exponential-logarithmic-equations-test-6":{"type":"choice","options":["$x=\\log_2 3,1$","$x=1$","$x=-3,2$"],"answer":[1],"choice_feedback":["Replacing the inadmissible root −3 by 3 changes the equation.","Set u = 2ˣ \u0026gt; 0. Then (u + 3)(u − 2) = 0; only u = 2 is possible, giving x = 1.","Those are roots of the substituted quadratic, not values of x."]}},"hints":[],"explanation":"Set u = 2ˣ \u0026gt; 0. Then (u + 3)(u − 2) = 0; only u = 2 is possible, giving x = 1.","id":"exponential-logarithmic-equations-test-6","coverage_topics":["aa-hl-exponential-equations"],"correct_feedback":"Set u = 2ˣ \u0026gt; 0. Then (u + 3)(u − 2) = 0; only u = 2 is possible, giving x = 1.","incorrect_feedback":"Set u = 2ˣ \u0026gt; 0. Then (u + 3)(u − 2) = 0; only u = 2 is possible, giving x = 1.","evidence_kind":"objective-concept-test"},{"content":"Which candidate is valid for $\\ln(x-4)+\\ln x=\\ln12$?\u003Cbr\u003E[[choice-exponential-logarithmic-equations-test-7]]","widgets":{"choice-exponential-logarithmic-equations-test-7":{"type":"choice","options":["6","-2","2"],"answer":[0],"choice_feedback":["The domain is x \u0026gt; 4; x² − 4x − 12 = 0 gives 6 or −2.","Both original arguments would be negative.","ln(x − 4) is undefined at 2, and 2 does not solve the quadratic."]}},"hints":[],"explanation":"The domain is x \u0026gt; 4; x² − 4x − 12 = 0 gives 6 or −2.","id":"exponential-logarithmic-equations-test-7","coverage_topics":["aa-hl-logarithmic-equations"],"correct_feedback":"The domain is x \u0026gt; 4; x² − 4x − 12 = 0 gives 6 or −2.","incorrect_feedback":"The domain is x \u0026gt; 4; x² − 4x − 12 = 0 gives 6 or −2.","evidence_kind":"objective-concept-test"},{"content":"Two quantities are $A(t)=30\\cdot2^t$ and $B(t)=120(1/2)^t$ for $t\\ge0$. Enter the time when they are equal. [[input-exponential-logarithmic-equations-test-8]]","widgets":{"input-exponential-logarithmic-equations-test-8":{"type":"input","answer":"1"}},"hints":[],"explanation":"Set the expressions equal: 2ᵗ = 4·2⁻ᵗ, so 2²ᵗ = 4 and t = 1.","id":"exponential-logarithmic-equations-test-8","coverage_topics":["aa-hl-exponential-equations"],"correct_feedback":"Set the expressions equal: 2ᵗ = 4·2⁻ᵗ, so 2²ᵗ = 4 and t = 1.","incorrect_feedback":"Set the two models equal and collect their powers of 2 on one side.","evidence_kind":"objective-concept-test"}],"mix":0,"tinkumo_module_mode":"test"}