how can I use logarithms to compare a,b,c in an ascending order $$a=\frac{3^{8.7} - 3^{6.2}}{5}\\\ b=\frac{3^{11.7} - 3^{8.7}}{6} \\\ c=\frac{3^{11.7} - 3^{6.2}}{11} \\\ $$ i tried to simplify a b and c to get a constant that won't effect the compression which was in this case $3^{6.2}$ and compared a and b then a and c then calculated both b and c to rank them from highest to lowest $$\\\ a=\frac{3^{6.2}(3^{2.5} - 1)}{5}\\\ b=\frac{3^{6.2}(3^{5.5} - 3^{2.5})}{6} \\\ c=\frac{3^{6.2}(3^{5.5} - 1)}{11} $$ but the lesson is about logarithms and when I added log to a b and c I could not really simplify it down
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Hint. (Without using logarithms.)
Compare, in ascending order, the slopes of these three secants
to the graph $y=3^x$
GEdgar
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1This is an excellent insight, +1. But does the OP know calculus, or even the difference quotient, if they are doing logarithms in school? It would be difficult for them to make the jump you did. – theREALyumdub Nov 24 '23 at 14:42
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I think i am starting to get it we use the difference quotient just as mentioned earlier to find the three slopes and use the slopes as a way to compare the magnitude yet when looking at the graph i found that the slopes in order are c>b>a but the result is b>c>a mind just clearing my confusion in the order of the slopes – lodo Nov 24 '23 at 17:09
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Your second version of the definitions makes clear that the term subtracted is rather small compared to the first term and the three expressions are not close enough for it to matter. You can also divide out the common factor $3^{6.2}$ as that will not change the ordering. Now you can use logs to compare the difference in the power of $3$ to the difference in the denominators to find the order.
Ross Millikan
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\textis not intended to use it that way. It's supposed to be used to insert little bits of text inside math expressions, like "is odd" or "continuous", not complete paragraphs. Instead of doing that, finish your first math expressions, add the text, and begin a new math expressions for the second list of equations. – jjagmath Nov 24 '23 at 14:19