Solving logarithmic equations?

hatjuice

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Joined
Apr 19, 2014
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6
From what I've seen in my textbook the process seems really straightforward. Yet when I actually try to solve one I always get the wrong solution (at least according to my calculator). I always get the same answer, but when I plug it in the answer doesn't work. I could really use some help...
Here's an example of what I have attempted to do. Any tips for solving them? Thanks in advance, I really appreciate it! c:
 

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From what I've seen in my textbook the process seems really straightforward. Yet when I actually try to solve one I always get the wrong solution (at least according to my calculator). I always get the same answer, but when I plug it in the answer doesn't work. I could really use some help...
Here's an example of what I have attempted to do. Any tips for solving them? Thanks in advance, I really appreciate it! c:


attachment.php


3 = log(8) + 3log(x)

3 = log(23)+ 3log(x)

3 = 3log(2)+ 3log(x)

1 = log(2) + log(x)

1 = log(2*x)

10 = 2*x

x = 5

Now check..

log(8) + 3log(x) = log(8) + 3log(5) = log(8) + log(53) = log(8*125) = log(1000) = 3 ............. checks
 
attachment.php


3 = log(8) + 3log(x)

3 = log(23)+ 3log(x)

3 = 3log(2)+ 3log(x)

1 = log(2) + log(x)

1 = log(2*x)

10 = 2*x

x = 5

Now check..

log(8) + 3log(x) = log(8) + 3log(5) = log(8) + log(53) = log(8*125) = log(1000) = 3 ............. checks
Thanks!
 
Your error was in going from \(\displaystyle log(8x^3)\) to \(\displaystyle 3 log(8x)\). That is incorrect because the exponent, 3, is not on both 8 and x. Since \(\displaystyle 8= 2^3\), you could have done this: \(\displaystyle log(8x^3)= log(2^3x^3)= log((2x)^3)= 3log(2x)= 3\). That is, of course, equivalent to what Subhotosh Khan wrote.
 
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