this question is trickier than i'm used too...a little help please? $\displaystyle log_{7}(49)(7) + log_{7}\sqrt{7}$
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Hi there. $\displaystyle log_7 49=2$ so $\displaystyle 7log_7 49=7\times2$ and $\displaystyle log_7\sqrt{7}=\frac{1}{2}$ So $\displaystyle 7log_7 49+log_7 \sqrt{7}=14\frac{1}{2}$
Originally Posted by Stroodle Hi there. $\displaystyle log_7 49=2$ so i don't understand this part, why did the 7 move to the front of the log?
Because it's in a separate bracket to the 49, it means that the logarithm is multiplied by 7. I just put it in front because that's how it is conventionally written (at least in my texts).
Originally Posted by Stroodle Hi there. $\displaystyle log_7 49=2$ so $\displaystyle 7log_7 49=7\times2$ and $\displaystyle log_7\sqrt{7}=\frac{1}{2}$ So $\displaystyle 7log_7 49+log_7 \sqrt{7}=14+\frac{1}{2}$ you forgot a sign.
Originally Posted by Raoh you forgot a sign. I don't get it? $\displaystyle 14+\frac{1}{2}=\frac{29}{2}=14\frac{1}{2}$
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