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log_(27)9=

log279= \log _{27} 9=

Full solution

Q. log279= \log _{27} 9=
  1. Find logarithm base and exponent: We need to find the value of log279\log_{27}9. This means we are looking for the exponent that 2727 must be raised to in order to get 99.
  2. Recognize powers of 33: Recognize that 2727 is a power of 33, specifically 27=3327 = 3^3, and 99 is also a power of 33, specifically 9=329 = 3^2.
  3. Rewrite logarithm in base 33: Rewrite the logarithm in terms of the base 33: log33(32)\log_{3^3}(3^2).
  4. Apply logarithm property: Use the property of logarithms that says logac(bd)=dcloga(b)\log_{a^c}(b^d) = \frac{d}{c} \cdot \log_a(b). In this case, log33(32)=23log3(3)\log_{3^3}(3^2) = \frac{2}{3} \cdot \log_3(3).
  5. Evaluate logarithm: Since log3(3)\log_3(3) is equal to 11 (because 33 raised to the power of 11 is 33), we have 23×1=23\frac{2}{3} \times 1 = \frac{2}{3}.
  6. Final result: Therefore, log279=23\log_{27}9 = \frac{2}{3}.