I know this is a very easy question, but somehow I just cannot solve it. Can someone please help me out. Thanks in advance.
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Hint: What is $2^3$? – Math1000 Nov 18 '19 at 03:43
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1@Math1000 Come on, that is not a hint. – Rushabh Mehta Nov 18 '19 at 03:43
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I meant log(x-7) – Kevin Nov 18 '19 at 03:48
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It is used in finding the answer, so I consider it a hint. Granted it may be too much of a hint... – Math1000 Nov 18 '19 at 03:49
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1Oh, you really did mean $\log_e(x-7)$. In that case the answer is the same, but the method based on rules of logarithms can no longer be used, because the base of the logarithms are different. In that case I see no way to solve the equation other than by inspection... – Math1000 Nov 18 '19 at 03:53
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Shouldn’t there be a more advanced way other than guess and check. The question really got me. I been applying all sorts of log formulas but non of them works. Is there a theorem that could work? – Kevin Nov 18 '19 at 03:56
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@Kevin There is a way. Note the change of base formula for logs. Convert the base into $\log_2$. – Rushabh Mehta Nov 18 '19 at 03:56
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3Possible duplicate of Intuition behind logarithm change of base – Rushabh Mehta Nov 18 '19 at 03:57
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Since $\log_a (1) = 0$, $x = 8$ gives $\log_2 (8) + \log(1) = 3$. Since the LHS is increasing, $x = 8$ must be the only solution. – KM101 Nov 18 '19 at 04:06
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@Don Thousand Could you elaborate on how I could apply that to this problem. I still got stuck on the point where I need to multiply two logs even if they have the same base. Thank you. – Kevin Nov 18 '19 at 04:52