How would i solve this? Log3^9-log3^27+log3^243
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Originally Posted by strdatmage How would i solve this? Log3^9-log3^27+log3^243 There is nothing there to solve. If you want to add those up then: $\displaystyle 9L+27L+243L=~?$ where $\displaystyle L=\log(3)$
Originally Posted by Plato There is nothing there to solve. If you want to add those up then: $\displaystyle 9L+27L+243L=~?$ where $\displaystyle L=\log(3)$ It says : Write each expression as a single logarithm. B. Find the value of each expression
Originally Posted by strdatmage It says : Write each expression as a single logarithm. B. Find the value of each expression That is exactly what I posted. Now you do some work for yourself.
Okay, so just Do that and replace L With 3?
Originally Posted by strdatmage It says : Write each expression as a single logarithm. B. Find the value of each expression As Plato demonstrated, you can use this property of logarithms: $\displaystyle \log_bx^a=a\log_bx,\quad x>0, b>0, b\ne1$. Then combine like terms. You should familiarize yourself with basic logarithmic identities. Consult your textbook for information.
Originally Posted by strdatmage How would i solve this? Log3^9-log3^27+log3^243 I'm a little bit confused about the writing of this sum ... Could it be that the original term was: $\displaystyle \log_3(9) - \log_3(27) + \log_3(243)$ If so the final result is 4.
Originally Posted by strdatmage It says : Write each expression as a single logarithm. B. Find the value of each expression 9L + 27L + 243L = 279L; where L = log(3) 279log(3) is the single logarithm Use your calculator to find the value the appropriate base as it wasn't detailed here (assuming base e or base 10)
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