How would I go about solving for "x" for something like this (should be fairly simple): $\displaystyle log7^x=log7^9$ edit: thats supposed to be base 7. how would you enter the code for that?
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Originally Posted by cjmac87 How would I go about solving for "x" for something like this (should be fairly simple): $\displaystyle log7^x=log7^9$ if log(something) = log(something) then the somethings must be equal, the log function is one-to-one. so just drop the logs. you have 7^x = 7^9, now what is x?
thanks that helps. ok what about something thats not one-to-one: $\displaystyle log_2{2^-1}=x$
Originally Posted by cjmac87 thanks that helps. ok what about something thats not one-to-one: $\displaystyle log_2{2^-1}=x$ you should know that $\displaystyle \log_aa^n = n$
ah, oh course. very simple. one last thing i need cleared up which is probably easy to: $\displaystyle log_2{16}$ can be solved how? how about $\displaystyle log_{16}{1/4}$?
Originally Posted by cjmac87 ah, oh course. very simple. one last thing i need cleared up which is probably easy to: $\displaystyle log_2{16}$ can be solved how? how about $\displaystyle log_{16}{1/4}$? $\displaystyle let \, \log_2 {16} = x$ then $\displaystyle 2^x = 16 = 2^4$ now, you can continue.. the other one, note that $\displaystyle \log_a {\frac{c}{d}} = \log_a {c} - \log_a {d}$ and $\displaystyle \, \log_a 1 = 0 \, for \, all \, a>0$ can you do the rest?
You seem to be missing an important concept: Logarithm = Exponent Trust me on this. What 2 roles are played by the 4, below? $\displaystyle log_{2}(16)\;=\;4\;\iff\;2^{4}\;=\;16$ You do the other one.
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