Question

Find all solutions of the following equation. $\log(x + 1) - \log(x + 4) - \log(4) = 0$ Check your answer(s) by plugging it/them into the equation. Write your list of values of $x$ between square brackets [ ] and ]. If there is more than one, separate them by commas; e.g. [-2, 4, 7]. If there is no solution, write empty (all lower case letters). Answer: $x = $

          Find all solutions of the following equation.
$\log(x + 1) - \log(x + 4) - \log(4) = 0$
Check your answer(s) by plugging it/them into the equation.
Write your list of values of $x$ between square brackets [ ] and ]. If there is more than one, separate them by commas; e.g. [-2, 4, 7].
If there is no solution, write empty (all lower case letters).
Answer: $x = $
        
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Find all solutions of the following equation.
log(x + 1) - log(x + 4) - log(4) = 0
Check your answer(s) by plugging it/them into the equation.
Write your list of values of x between square brackets [ ] and ]. If there is more than one, separate them by commas; e.g. [-2, 4, 7].
If there is no solution, write empty (all lower case letters).
Answer: x =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find all solutions of the following equation. log(x+1)-log(x+4)-log(4)=0 Check your answer(s) by plugging it/them into the equation. Write your list of values of x between square brackets [ and ]. If there is more than one, separate them by commas; e.g. -2,4,7. If there is no solution, write empty (all lower case letters). Answer: x= Find all solutions of the following equation 1ogx+1-logx+4-log4=0 Check your answer(s) by plugging it/them into the equation. Write your list of values of between square brackets [and]. If there is more than one, separate them by commas;e.g. [-2,4,7] If there is no solution, write empty all lower case letters Answer:=
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Transcript

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00:01 When finding solutions to equations, there are two things that we want to look at.
00:04 The first one is going to be the denominator.
00:07 So we know in an equation like this, the denominator cannot be zero, otherwise the solution would be undefined.
00:15 So we want to find all defined solutions.
00:18 Therefore, we know that the denominator cannot equal zero.
00:23 The second part is the numerator.
00:25 We want to look at is there anything in the numerator that could potentially create an undefined solution.
00:31 In this example, no.
00:33 X can be any number and the solution will still be defined...
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