Solving Exponential Equations with Logarithms - Algebra 2

  1. Rewrite the solution of a⋅10ct=da\cdot 10^{ct}=d using ln⁡\ln only.
    t=ln⁡(da)cln⁡(10)t=\frac{\ln\left(\frac{d}{a}\right)}{c\ln(10)}. Apply change-of-base formula: log⁡10(x)=ln⁡(x)ln⁡(10)\log_{10}(x) = \frac{\ln(x)}{\ln(10)}.
  2. State the change-of-base formula for rewriting log⁡b(x)\log_b(x) using log⁡\log.
    log⁡b(x)=log⁡(x)log⁡(b)\log_b(x)=\frac{\log(x)}{\log(b)}. Converts any logarithm base to common logarithm form.
  3. Find tt in 103t=101210^{3t}=10^{12} and give the exact value.
    t=4t=4. Since the bases are equal, equate exponents: 3t=123t = 12, so t=4t = 4.
  4. Find tt in 4⋅2t=644\cdot 2^{t}=64 and express the result as a logarithm.
    t=log⁡2(16)t=\log_2(16). Divide by 4: 2t=162^t = 16, then take base-2 logarithm.
  5. Find tt in 0.2⋅104t=20.2\cdot 10^{4t}=2 and express the result as a logarithm.
    t=log⁡(10)4t=\frac{\log(10)}{4}. Multiply by 5: 104t=1010^{4t} = 10, then take log and divide by 4.
  6. Find tt in 5⋅102t=5005\cdot 10^{2t}=500 and give the exact value.
    t=1t=1. Since log⁡(100)=2\log(100) = 2 and we divide by 2, we get t=1t = 1.
  7. Identify the final step to isolate tt from ct=log⁡b(da)ct=\log_b\left(\frac{d}{a}\right).
    Divide by cc: t=log⁡b(da)ct=\frac{\log_b\left(\frac{d}{a}\right)}{c}. Divide both sides by coefficient cc to solve for tt.
  8. Find tt in 2⋅et=102\cdot e^{t}=10 and express the result as a logarithm.
    t=ln⁡(5)t=\ln(5). Divide by 2: et=5e^t = 5, then take natural logarithm.
  9. What is the solution for tt in a⋅10ct=da\cdot 10^{ct}=d written using common logarithm?
    t=log⁡(da)ct=\frac{\log\left(\frac{d}{a}\right)}{c}. Divide by aa to get 10ct=da10^{ct} = \frac{d}{a}, then apply common log.
  10. Find tt in 2⋅103t=502\cdot 10^{3t}=50 and express the result as a logarithm.
    t=log⁡(25)3t=\frac{\log(25)}{3}. Divide by 2: 103t=2510^{3t} = 25, then take log and divide by 3.
  11. Identify the correct log form of the solution to a⋅ect=da\cdot e^{ct}=d.
    t=ln⁡(da)ct=\frac{\ln\left(\frac{d}{a}\right)}{c}. Standard form using natural logarithm for base ee exponentials.
  12. Rewrite the solution of a⋅2ct=da\cdot 2^{ct}=d using ln⁡\ln only.
    t=ln⁡(da)cln⁡(2)t=\frac{\ln\left(\frac{d}{a}\right)}{c\ln(2)}. Apply change-of-base formula: log⁡2(x)=ln⁡(x)ln⁡(2)\log_2(x) = \frac{\ln(x)}{\ln(2)}.
  13. Find tt in 0.1⋅10t=10.1\cdot 10^{t}=1 and express the result as a logarithm.
    t=log⁡(10)t=\log(10). Multiply by 10: 10t=1010^t = 10, so t=log⁡(10)t = \log(10).
  14. Find tt in 0.5⋅et=40.5\cdot e^{t}=4 and express the result as a logarithm.
    t=ln⁡(8)t=\ln(8). Multiply by 2: et=8e^t = 8, then take natural logarithm.
  15. What property lets you move an exponent out front: log⁡b(xk)=?\log_b(x^k)=?
    log⁡b(xk)=klog⁡b(x)\log_b(x^k)=k\log_b(x). Uses the power property to bring exponents in front.
  16. Find tt in 1.5⋅20.5t=61.5\cdot 2^{0.5t}=6 and express the result as a logarithm.
    t=log⁡2(4)0.5t=\frac{\log_2(4)}{0.5}. Divide by 1.5: 20.5t=42^{0.5t} = 4, then take log and divide by 0.5.
  17. Find tt in 6⋅2t=36\cdot 2^{t}=3 and express the result as a logarithm.
    t=log⁡2(12)t=\log_2\left(\frac{1}{2}\right). Divide by 6: 2t=122^t = \frac{1}{2}, then take base-2 logarithm.
  18. Find tt in 3⋅2t=243\cdot 2^{t}=24 and give the exact value.
    t=3t=3. Since log⁡2(8)=3\log_2(8) = 3, we get t=3t = 3.
  19. Identify the inverse statement that justifies taking logs: by=xb^{y}=x implies what?
    y=log⁡b(x)y=\log_b(x). Logarithm is the inverse function of exponentiation.
  20. What is the solution for tt in a⋅ect=da\cdot e^{ct}=d written using natural logarithm?
    t=ln⁡(da)ct=\frac{\ln\left(\frac{d}{a}\right)}{c}. Divide by aa to get ect=dae^{ct} = \frac{d}{a}, then apply natural log.
  21. Find tt in 0.01⋅10t=1000.01\cdot 10^{t}=100 and express the result as a logarithm.
    t=log⁡(10000)t=\log(10000). Multiply by 100: 10t=1000010^t = 10000, so t=log⁡(10000)t = \log(10000).
  22. Identify the correct log form of the solution to a⋅2ct=da\cdot 2^{ct}=d.
    t=log⁡2(da)ct=\frac{\log_2\left(\frac{d}{a}\right)}{c}. Standard form using base-2 logarithm for base 2 exponentials.
  23. Find tt in 5⋅e2t=405\cdot e^{2t}=40 and express the result as a logarithm.
    t=ln⁡(8)2t=\frac{\ln(8)}{2}. Divide by 5: e2t=8e^{2t} = 8, then take ln and divide by 2.
  24. Find tt in 3⋅22t=963\cdot 2^{2t}=96 and express the result as a logarithm.
    t=log⁡2(32)2t=\frac{\log_2(32)}{2}. Divide by 3: 22t=322^{2t} = 32, then take log and divide by 2.
  25. State the change-of-base formula for rewriting log⁡b(x)\log_b(x) using ln⁡\ln.
    log⁡b(x)=ln⁡(x)ln⁡(b)\log_b(x)=\frac{\ln(x)}{\ln(b)}. Converts any logarithm base to natural logarithm form.
  26. Find tt in e2t=e6e^{2t}=e^{6} and give the exact value.
    t=3t=3. Since the bases are equal, equate exponents: 2t=62t = 6, so t=3t = 3.
  27. Find tt in 7⋅e3t=17\cdot e^{3t}=1 and express the result as a logarithm.
    t=ln⁡(17)3t=\frac{\ln\left(\frac{1}{7}\right)}{3}. Divide by 7: e3t=17e^{3t} = \frac{1}{7}, then take ln and divide by 3.
  28. Find tt in 9⋅23t=729\cdot 2^{3t}=72 and express the result as a logarithm.
    t=log⁡2(8)3t=\frac{\log_2(8)}{3}. Divide by 9: 23t=82^{3t} = 8, then take log and divide by 3.
  29. What base restriction is required for log⁡b(x)\log_b(x) when solving abct=da b^{ct}=d?
    b>0b>0 and b≠1b\ne 1. Base must be positive and not equal to 1 for valid logarithm.
  30. Find tt in 3⋅e0.1t=93\cdot e^{0.1t}=9 and express the result as a logarithm.
    t=ln⁡(3)0.1t=\frac{\ln(3)}{0.1}. Divide by 3: e0.1t=3e^{0.1t} = 3, then take ln and divide by 0.1.
  31. Find tt in 8⋅10−t=28\cdot 10^{-t}=2 and express the result as a logarithm.
    t=log⁡(14)−1t=\frac{\log\left(\frac{1}{4}\right)}{-1}. Divide by 8: 10−t=1410^{-t} = \frac{1}{4}, then take log and divide by -1.
  32. What is the solution for tt in a⋅2ct=da\cdot 2^{ct}=d written using base-22 logarithm?
    t=log⁡2(da)ct=\frac{\log_2\left(\frac{d}{a}\right)}{c}. Divide by aa to get 2ct=da2^{ct} = \frac{d}{a}, then apply base-2 log.
  33. Find tt in 3⋅102t=3003\cdot 10^{2t}=300 and express the result as a logarithm.
    t=log⁡(100)2t=\frac{\log(100)}{2}. Divide by 3: 102t=10010^{2t} = 100, then take log and divide by 2.
  34. Find tt in 7⋅10t=1.47\cdot 10^{t}=1.4 and express the result as a logarithm.
    t=log⁡(15)t=\log\left(\frac{1}{5}\right). Divide by 7: 10t=1.47=0.2=1510^t = \frac{1.4}{7} = 0.2 = \frac{1}{5}, then take log.
  35. Find tt in 7⋅et=77\cdot e^{t}=7 and give the exact value.
    t=0t=0. Since ln⁡(1)=0\ln(1) = 0 when et=1e^t = 1, we get t=0t = 0.
  36. Rewrite the solution of a⋅ect=da\cdot e^{ct}=d using log⁡\log only.
    t=log⁡(da)clog⁡(e)t=\frac{\log\left(\frac{d}{a}\right)}{c\log(e)}. Apply change-of-base formula: ln⁡(x)=log⁡(x)log⁡(e)\ln(x) = \frac{\log(x)}{\log(e)}.
  37. Find tt in 4⋅10t=404\cdot 10^{t}=40 and give the exact value.
    t=1t=1. Since log⁡(10)=1\log(10) = 1, we get t=1t = 1.
  38. Find tt in 2⋅2t=642\cdot 2^{t}=64 and give the exact value.
    t=5t=5. Since log⁡2(32)=5\log_2(32) = 5, we get t=5t = 5.
  39. Identify the correct log form of the solution to a⋅10ct=da\cdot 10^{ct}=d.
    t=log⁡(da)ct=\frac{\log\left(\frac{d}{a}\right)}{c}. Standard form using common logarithm for base 10 exponentials.
  40. Find tt in 5⋅10t=2005\cdot 10^{t}=200 and express the result as a logarithm.
    t=log⁡(40)t=\log(40). Divide by 5: 10t=4010^t = 40, then take common logarithm.
  41. What property lets you rewrite log⁡b(da)\log_b\left(\frac{d}{a}\right) as a difference of logs?
    log⁡b(da)=log⁡b(d)−log⁡b(a)\log_b\left(\frac{d}{a}\right)=\log_b(d)-\log_b(a). Uses the quotient property: log⁡b(xy)=log⁡b(x)−log⁡b(y)\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y).
  42. What is the domain requirement for log⁡b(x)\log_b(x) in solving exponential equations?
    x>0x>0. Logarithm is only defined for positive arguments.
  43. Find tt in 24t=2202^{4t}=2^{20} and give the exact value.
    t=5t=5. Since the bases are equal, equate exponents: 4t=204t = 20, so t=5t = 5.
  44. Find tt in 0.25⋅22t=80.25\cdot 2^{2t}=8 and express the result as a logarithm.
    t=log⁡2(32)2t=\frac{\log_2(32)}{2}. Multiply by 4: 22t=322^{2t} = 32, then take log and divide by 2.
  45. Find tt in 0.5⋅2t=10.5\cdot 2^{t}=1 and express the result as a logarithm.
    t=log⁡2(2)t=\log_2(2). Multiply by 2: 2t=22^t = 2, so t=log⁡2(2)t = \log_2(2).
  46. Find tt in 2⋅e−t=82\cdot e^{-t}=8 and express the result as a logarithm.
    t=ln⁡(4)−1t=\frac{\ln(4)}{-1}. Divide by 2: e−t=4e^{-t} = 4, then take ln and divide by -1.
  47. Find tt in 10⋅e−0.2t=510\cdot e^{-0.2t}=5 and express the result as a logarithm.
    t=ln⁡(12)−0.2t=\frac{\ln\left(\frac{1}{2}\right)}{-0.2}. Divide by 10: e−0.2t=12e^{-0.2t} = \frac{1}{2}, then take ln and divide by -0.2.
  48. What log step isolates ctct after dividing: bct=dab^{ct}=\frac{d}{a}?
    ct=log⁡b(da)ct=\log_b\left(\frac{d}{a}\right). Take log⁡b\log_b of both sides to eliminate the exponential.
  49. Find tt in 9⋅2−t=729\cdot 2^{-t}=72 and express the result as a logarithm.
    t=log⁡2(8)−1t=\frac{\log_2(8)}{-1}. Divide by 9: 2−t=82^{-t} = 8, then take log and divide by -1.
  50. Find tt in 16⋅2t=116\cdot 2^{t}=1 and express the result as a logarithm.
    t=log⁡2(116)t=\log_2\left(\frac{1}{16}\right). Divide by 16: 2t=1162^t = \frac{1}{16}, then take base-2 log.
  51. Identify the first algebra step to solve abct=da b^{ct}=d for tt.
    Divide: bct=dab^{ct}=\frac{d}{a}. Isolate the exponential term before taking logarithms.
  52. Find tt in 12⋅100.5t=312\cdot 10^{0.5t}=3 and express the result as a logarithm.
    t=log⁡(14)0.5t=\frac{\log\left(\frac{1}{4}\right)}{0.5}. Divide by 12: 100.5t=1410^{0.5t} = \frac{1}{4}, then take log and divide by 0.5.
  53. State the solution for tt in abct=da b^{ct}=d written using a logarithm.
    t=log⁡b(da)ct=\frac{\log_b\left(\frac{d}{a}\right)}{c}. Isolate bctb^{ct} by dividing by aa, then take log⁡b\log_b of both sides.

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