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9log3209^{\log_{3}20}=

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Q. 9log3209^{\log_{3}20}=
  1. Understand the expression: Understand the expression 9log3209^{\log_{3}20}. We have an exponentiation where the base is 99 and the exponent is the logarithm of 2020 with base 33.
  2. Express base in terms: Express the base 99 in terms of base 33.\newlineSince 99 is 33 squared (323^2), we can rewrite the base as 323^2.
  3. Substitute base in expression: Substitute the expression for 99 into the original expression.\newlineNow we have (32)(log320)(3^2)^{(\log_{3}20)}.
  4. Apply power rule for exponents: Apply the power rule for exponents amn=(am)na^{m*n} = (a^m)^n. We can rewrite the expression as 32log3203^{2*\log_{3}20}.
  5. Use logarithm power rule: Use the logarithm power rule.\newlineThe power rule states that alogb(c)=logb(ca)a\log_b(c) = \log_b(c^a). Therefore, we can rewrite the expression as 3log3(202)3^{\log_{3}(20^2)}.
  6. Simplify using property of logarithms: Simplify the expression using the property of logarithms that blogb(x)=xb^{\log_b(x)} = x. Since the base of the logarithm and the base of the exponentiation are the same (3)(3), the expression simplifies to just 20220^2.
  7. Calculate 20220^2: Calculate 20220^2.\newline20220^2 is 400400.

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