Can someone please help me start on this: thanks
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Originally Posted by bolsius Can someone please help me start on this: thanks Broken link CB
Originally Posted by bolsius Can someone please help me start on this: thanks lol. Can't see anything. But while I'm here I may as well throw down some properties of logs in case they help... . as long as . where [tex]e[/math[ is the exponential function.
Originally Posted by Deadstar lol. Can't see anything. But while I'm here I may as well throw down some properties of logs in case they help... . as long as . where [tex]e[/math[ is the exponential function. All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here ln (1-x) + ln (1+x) = ln (3/4) thanks
Originally Posted by bolsius All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here ln (1-x) + ln (1+x) = ln (3/4) thanks Use the laws of logarithms to combine the two logs on the left: can you take it from there? CB
Originally Posted by bolsius All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here ln (1-x) + ln (1+x) = ln (3/4) thanks (remember this rule). So you have... Solve for x. EDIT: too slow...
Yes thanks I got it...I wasnīt sure if I had to use the laws of logarithms both side of the equal sign (right and left)
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