Solve the logarithmic equation. log ( x + 2 ) + log 2 = 3 \log\left(x+2\right)+\log2=3 lo g ( x + 2 ) + lo g 2 = 3

In this problem, we're asked to solve the log equation. And the equation we're given is log of x+2+2 is equal to 3. Now because I have this constant 3 here, I know that I can't just have 2 logs with the same base that I'm going to be able to just set things equal to each other, so I know that I'm going to have to use my steps here. Now the first thing I want to do is to isolate my log expression. Now looking at this since I have these 2 logs here, I wanna go ahead and condense this down into 1 log to make this a little bit more simple. Now these have the same base and they're being added together, so that tells me that I can go ahead and use the product rule here. So using the product rule, this becomes a log multiplying these together, 2 times x plus 2. Now we can actually further simplify this down by distributing this 2 in. So this becomes a log of 2x+4. And that's equal to that number that I have on the other side, 3. Now that I've isolated my log expression, I just have one log expression by itself, I can go ahead and move on to step 2 and put this into exponential form. Now, I know that this log is just the common log, which means that it's really a log base 10. So starting with our base here to put this in exponential form, starting with that base, raising it to the power of what's on the other side of the equal sign, so 10 to the power of 3, and then coming back around to that other side of the equal sign, 10 to the power of 3 is equal to 2x+4. Now we're in that exponential form. We've completed step 2, and we can go ahead and just solve for x. Now this 10 cubed I can go ahead and pull this out to simplify it because 10 cubed is just 10 times 10 times 10 which I know is just a 1,000. So this is really a 1000 is equal to 2x+4. Now to solve for x here, I want to go ahead and subtract 4 from both sides, canceling over here, leaving me with 996 is equal to 2x. Now one final step in isolating x is to divide both sides by 2, canceling it over here, and leaving me with my final answer that x is equal to 498. So this is my final answer but I do want to perform one final check here to make sure that this actually is my solution by plugging it back into the original argument of my log or what I was taking the log of. So I was originally taking the log of 2x+4. So I want to plug that 498 back into it in order to check. So this is 2 times 498 plus 4. And 2 times 498 is that 996 that we originally had. And then a +4 actually just takes us back to 1,000, which is greater than a 0, so we are done. And this is our final answer, x is equal to 498. Let me know if you have any questions.

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6. Exponential and Logarithmic Functions

Solving Exponential and Logarithmic Equations

Precalculus

6. Exponential and Logarithmic Functions

Solving Exponential and Logarithmic Equations

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Multiple Choice

Solve the logarithmic equation.
$\log\left(x+2\right)+\log2=3$

498

1998

No Solution

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Verified step by step guidance

Use the property of logarithms that states $ \log_a b + \log_a c = \log_a (bc) $ to combine the logarithms on the left side of the equation: $ \log((x+2) \cdot 2) = 3 $.

Simplify the expression inside the logarithm: $ \log(2(x+2)) = 3 $.

Rewrite the equation in exponential form to eliminate the logarithm: $ 2(x+2) = 10^3 $.

Solve for $ x $ by first expanding the left side: $ 2x + 4 = 1000 $.

Isolate $ x $ by subtracting 4 from both sides and then dividing by 2: $ 2x = 996 $, so $ x = 498 $.