Log B (A^C ) = would the right answer be? B log C A
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Oct 31, 2005 #2 Hi, I'm not exactly sure what you're trying to do but if you were to apply the log rule: \(\displaystyle log_a{x^n} = nlog_a{x}\) then you would get \(\displaystyle C\log_B{A}\) If you wanted to change the log base to C then you're off a bit... So if you could clarify what the goal is that would be great.
Hi, I'm not exactly sure what you're trying to do but if you were to apply the log rule: \(\displaystyle log_a{x^n} = nlog_a{x}\) then you would get \(\displaystyle C\log_B{A}\) If you wanted to change the log base to C then you're off a bit... So if you could clarify what the goal is that would be great.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Oct 31, 2005 #3 Hello, adam40g! \(\displaystyle \log_{_B}(A^C ) =\) Would the right answer be: \(\displaystyle B\log_{_C}(A)\) . . . . no Click to expand... The "third" property of logs is: . \(\displaystyle \log_b(X^n) \;= \;n\cdot\log_b(X)\) . . In baby-talk: the exponent "comes down in front".
Hello, adam40g! \(\displaystyle \log_{_B}(A^C ) =\) Would the right answer be: \(\displaystyle B\log_{_C}(A)\) . . . . no Click to expand... The "third" property of logs is: . \(\displaystyle \log_b(X^n) \;= \;n\cdot\log_b(X)\) . . In baby-talk: the exponent "comes down in front".
