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Algebra · real student question

Solve 32log3(x - 5) = 49 for x.

Question

Solve

32log⁡3(x−5)=4932\log_3(x-5)=49

for xx.

Step-by-step solution

  1. State the domain before solving. A logarithm needs a strictly positive argument, so

    x−5>0⟹x>5x-5>0\qquad\Longrightarrow\qquad x>5

    Any candidate answer must clear this bar; writing the restriction down first is what makes the final check meaningful rather than an afterthought.

  2. Isolate the logarithm. Divide both sides by the coefficient 3232:

    log⁡3(x−5)=4932\log_3(x-5)=\frac{49}{32}

    Leaving the right side as the exact fraction 4932=1.53125\tfrac{49}{32}=1.53125 keeps the answer exact.

  3. Convert to exponential form. The defining equivalence for logarithms is

    log⁡b(A)=c  ⟺  A=b c\log_b(A)=c\iff A=b^{\,c}

    Here b=3b=3, A=x−5A=x-5, c=4932c=\tfrac{49}{32}, so

    x−5=349/32x-5=3^{49/32}

  4. Solve for x. Add 55:

    x=5+349/32x=5+3^{49/32}

    This is the exact answer; no further simplification is possible because 4932\tfrac{49}{32} is not an integer and 4949 and 3232 share no factor with 33.

  5. Evaluate numerically and confirm the domain. Since 349/32=31.53125=5.377643^{49/32}=3^{1.53125}=5.37764,

    x=5+5.37764=10.37764x=5+5.37764=10.37764

    As a size check, 31.5=27≈5.1963^{1.5}=\sqrt{27}\approx5.196 and 31.531253^{1.53125} should be a little larger — it is ✓. And x≈10.378>5x\approx10.378>5, so the logarithm is defined and the solution is valid.

  6. Verify by substituting back. With x=10.3776424x=10.3776424: x−5=5.3776424x-5=5.3776424, and log⁡3(5.3776424)=ln⁡5.3776424ln⁡3=1.531250\log_3(5.3776424)=\dfrac{\ln 5.3776424}{\ln 3}=1.531250, so 32×1.53125=49.000032\times1.53125=49.0000 ✓.

Answer

x=5+349/32≈10.3776x=5+3^{49/32}\approx 10.3776

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