Help with these two would be grealy appreciated.
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Originally Posted by Fnus Help with these two would be grealy appreciated. 2. From equation (2): . Substitute this into equation (1): .......
Originally Posted by Fnus Help with these two would be grealy appreciated. 1. On the left hand side, you're missing an argument for the log. Nevertheless: Note that and so the right hand side is . The left hand side can also be expressed as a single log. Do it. Now equate the arguments of the logs on each side and solve for x.
Yay, I get y = 1, and x = 3, and I'm pretty sure that checks out. Thanks a ton. (:
Oh, yeah, on the left side I'm missing an x. But I don't get what it is you're trying to do by
Originally Posted by Fnus Oh, yeah, on the left side I'm missing an x. But I don't get what it is you're trying to do by Then the left hand side can be written . The right hand side can be written . Therefore Warning: Although the quadratic equation has two solutions, only one is valid (why?).
Oh *okay*. And I get x = -5 or 10, but it can't be -5 because you can't take the logarith of a negative number, I think..
Originally Posted by Fnus Oh *okay*. And I get x = -5 or 10, but it can't be -5 because you can't take the logarith of a negative number, I think..
Now we note from here the bases are the same so the exponents must be equal. So we get now we multiply it by -2 and add to the one below to get Subbing back in we get
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