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And this is really straight derived from these two right over here.
We're asked to solve the log of x plus log of 3 is equal to 2 log of 4 minus log of 2. So we have the log of x plus the log of 3 is equal to 2 times the log of 4 minus the log of 2, or the logarithm of 2. Whenever you see a logarithm written without a base, the implicit base is 10.
So we could write 10 here, 10 here, 10 here, and 10 here.
Round-off error can get really big really fast with logs, and you don't want to lose points because you rounded too early and thus too much. In other words, when you plug your decimal approximation into the original equation, you're just making sure that the result is close enough to be reasonable.
For instance, to check the solution of the equation No, the two values are not equal, but they're pretty darned close.
An example would be: , is a valid solution, and often this will be all that I'm supposed to give for the answer.
However, in this case (maybe leading up to graphing or word problems) they want me to provide a decimal approximation.
This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. And 10 to the same power is going to be equal to 8. 3x is equal to 8, and then we can divide both sides by 3.
So the right-hand side simplifies to log base 10 of 8. Divide both sides by 3, you get x is equal to 8 over 3.
Allowing for round-off error, these values confirm to me that I've gotten the right answer.
If, on the other hand, my solution had returned a value of, say, Let's do a couple more examples. is the answer, I first need to check (especially because this answer is negative) whether it'll work in the original equation.