(x/5)^log_b(5) - (x/6)^log_b(6) = 0
this post was submitted on 15 May 2024
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Expanding on my reply, here's a different way to continue your own work that would have yielded both solutions, by avoiding any steps that divide by x:
solution
Starting from the third line of your work:(x/5)^ln(5) = (x/6)^ln(6)
x^ln(5) / 5^ln(5) = x^ln(6) / 6^ln(6) → Distribute exponents
x^ln(5) * 6^ln(6) - x^ln(6) * 5^ln(5) = 0 → Cross multiply, move terms to one side
x^ln(5) * (6^ln(6) - x^ln(1.2) * 5^ln(5)) = 0 → Factor out x^ln(5)
Can set each factor to 0:
x^ln(5) = 0 yields x = 0
6^ln(6) - x^ln(1.2) * 5^ln(5) = 0
6^ln(6) / 5^ln(5) = x^ln(1.2) → Add right term to right side, divide by its coefficient
(5^(ln(6)/ln(5)))^ln(6) / 5^ln(5) = x^ln(1.2) → Convert numerator of left side to have same base as denominator, using change of base formula: log_5(6) = ln(6)/ln(5)
5^(ln(6)^2 / ln(5)) / 5^ln(5) = x^ln(1.2) → Simplify exponent of numerator slightly
5^((ln(6)^2 - ln(5)^2) / ln(5)) = x^ln(1.2) → Combine terms on left side, simplify numerator into a single fraction
5^((ln(6)+ln(5))(ln(6)-ln(5))/ln(5)) = x^ln(1.2) → Factor exponent numerator as difference of squares
5^(ln(30)ln(1.2)/ln(5)) = x^ln(1.2) → Simplify sum and difference of logs in left exponent numerator
30^ln(1.2) = x^ln(1.2) → By change of base formula again, ln(30)/ln(5) = log_5(30), so 5^(ln(30)/ln(5)) = 30
x = 30