Someone help to explain this please!!
log x + log (x + 48) = 2

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- anonymous

Someone help to explain this please!!
log x + log (x + 48) = 2

- jamiebookeater

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- campbell_st

well using log laws for multiplication the problem can be written as
\[\log(x(x + 48)) = 2~~~or~~~~\log(x^2 + 48x) = 2\]
if this is a base 10 log system them raise each side of the equation to a power of 10
and you need to know the log law
\[a^{\log_{a}(b)} = b\]
so
\[10^{\log(x^2 + 48x)} = 10^2\]
which becomes
\[x^2 + 48x = 100\]
now you can solve for x
and remember you can't take the log of a negative number when you consider the solutions.

- Hero

@calecea Any questions about Campbell_st approach here?

- anonymous

@Hero no, I think I got it.

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- anonymous

@campbell_st , so does x=4.9? I did square root to both sides, then added the x and got 49, then divided by 10.

- campbell_st

well I would rewrite it as
\[x^2 + 48x - 100 = 0\]
and this factors to
(x - 2)(x + 50) = 0
so the solutions seem to be
x = 2 and x = -50

- anonymous

oh, I didn't know it was supposed to be written that way.

- campbell_st

well lets look at the positive value x = 2
you get
log(2) + log(2 + 48)
this can be rewritten as
log(2 x 50) = log(100)
and 100 = 10^2
so it seems to make sense

- anonymous

oh ok, Thank you!

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