Logarithm help needed

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dttrapo

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Feb 14, 2021
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Working on logarithms, and I am absolutely stumped by this one:

Solve for x:

log(x)+log(x+5)=4

Answer rounded to 4 decimal places.

I believe it can be re-written to log(x)(x+5)=4 and then simplified to log(x^2+5x)=4 and from there I honestly have no idea where to go. Any help explaining the steps to solve would be greatly appreciated!
 
dttrapo said:
Working on logarithms, and I am absolutely stumped by this one:

Solve for x:

log(x)+log(x+5)=4

Answer rounded to 4 decimal places.

I believe it can be re-written to log(x)(x+5)=4 and then simplified to log(x^2+5x)=4 and from there I honestly have no idea where to go. Any help explaining the steps to solve would be greatly appreciated!
Click to expand...
Use:

loga(b) = c → b = ac
 
The point of Subhotosh’s hint is this.

If you had [MATH]log_{10}(x^2 + 5x) = log_{10}(p)[/MATH]
then you could say [MATH]x^2 + 5x = p[/MATH]. WHY?

Then you would have a quadratic.

So can you turn that 4 into some log to the base 10?
 
JeffM said:
The point of Subhotosh’s hint is this.

If you had [MATH]log_{10}(x^2 + 5x) = log_{10}(p)[/MATH]
then you could say [MATH]x^2 + 5x = p[/MATH]. WHY?

Then you would have a quadratic.

So can you turn that 4 into some log to the base 10?
Click to expand...

So it would became x^2+5x-10000=0 And solve as a quadratic? I'm trying to figure out what on earth could be the factors for this one. Thank you both for the hints so far, I'm really struggling with this and every little bit helps.
 
dttrapo said:
So it would became x^2+5x-10000=0 And solve as a quadratic? I'm trying to figure out what on earth could be the factors for this one. Thank you both for the hints so far, I'm really struggling with this and every little bit helps.
Click to expand...
Do you the solutions of roots of quadratic equations?
 
If it's taking too long to find the factors or if they are not "nice", you can always use the quadratic formula.
The fact that the question says "Answer rounded to 4 decimal places" is an indication that you won't be able to easily find the factors.

Also, when you get your 2 solutions to the quadratic equation, check them back in the original log equation. (One may not be valid.)
 
dttrapo said:
So it would became x^2+5x-10000=0 And solve as a quadratic? I'm trying to figure out what on earth could be the factors for this one. Thank you both for the hints so far, I'm really struggling with this and every little bit helps.
Click to expand...
That is indeed the equation you need to solve.

I always say: if you cannot factor a quadratic in 30 seconds, use the quadratic formula.

And always pay attention to cats: they are descended from the gods. Can you take the log of a negative number?
 
Please watch this video to see a better method to solving quadratics than using the quadratic equation. It is the method by Po-Shen Loh.
 
I like the method, but in your third example in the video, you get x = - 2 by your proposed method and x = 2 by the quadratic formula. Start watching about minute 11.
 
JeffM said:
I like the method, but in your third example in the video, you get x = - 2 by your proposed method and x = 2 by the quadratic formula. Start watching about minute 11.
Click to expand...
Yeah, I noticed that a few minutes ago as well. I need to redo that video. Thanks for letting me know!
 
It's a nice video, but the "new" method boils down to doing this...

[math] \frac{-b}{2}\pm{\sqrt{\frac{b^2}{4}-c}} [/math]
...which is just the quadratic formula with a=1, and written in a slightly different form.

However, I guess this approach might suit people who are better at memorising methods rather than memorising formulas.
 
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