Logarithm Equation

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Vertciel

Junior Member
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May 13, 2007
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Hello there,

I am new to logarithms and am studying them by myself. Could anyone please give a hint (not the answer please!) as to how to solve for x in this equation?

-3 log (x + 2) + log 5 = 0

The answer is here in white: x = -2 + 5^(1/3)

Thank you very much![/tex]
 
Hint: You know that you can add two things together to get zero, and so what you want to do in order to get to a point where you can solve for x without logarithms being involved is find some sort of equivalency, such as moving one of the logarithms to the other side (add or subtract from both sides- I forget what the sign was on it, but I am sure you can cope).

- Bryan
 
You could rewrite as:

\(\displaystyle \L\\log(\frac{5}{(x+2)^{3}})=0\)

\(\displaystyle \L\\\frac{5}{(x+2)^{3}}=10^{0}=1\)

So, you have: \(\displaystyle \L\\\frac{5}{(x+2)^{3}}=1\)

Now, solve for x and you will get your answer.
 
galactus said:
You could rewrite as:

\(\displaystyle \L\\log(\frac{5}{(x+2)^{3}})=0\)
Click to expand...

Hello there,

Thank you very much, kamzure and galactus, for replying.

Galactus, could you please show me how you got to your first step?

Also, would it be possible to equate the two logarithms since they are both in base 10?

-3 log (x + 2) + log 5 = 0

log (x + 2)^-3 + log 5 = 0

(x + 2)^3 = 5

Thanks again.
 
Vertciel said:
Also, would it be possible to equate the two logarithms since they are both in base 10?
Click to expand...
No! That is not the point.

This is the basic idea: \(\displaystyle \log _b (t) = 0\quad \Rightarrow \quad t = 1\) for any positive base b.
 
Vertciel said:
galactus said:
You could rewrite as:

\(\displaystyle \L\\log(\frac{5}{(x+2)^{3}})=0\)
Click to expand...

Hello there,

Thank you very much, kamzure and galactus, for replying.

Galactus, could you please show me how you got to your first step?

Also, would it be possible to equate the two logarithms since they are both in base 10?

-3 log (x + 2) + log 5 = 0

log (x + 2)^-3 + log 5 = 0

(x + 2)^3 = 5



Thanks again.
Click to expand...

You need to use the rules for logs:

a log b = log b<SUP>a</SUP>

So,

-3 log(x + 2) + 5 = 0

becomes

log (x + 2)<SUP>-3</SUP> + log 5 = 0

Now, another rule of logs:

log a + log b = log (a*b)

So,

log (x + 2)<SUP>-3</SUP> + log 5 = 0

becomes

log [5 *(x + 2)<SUP>-3</SUP>) ] = 0

Now, as you've already been told, change this back to exponential form:

if log a = b, then a = 10^b

5*(x + 2)<SUP>-3</SUP> = 10<SUP>0</SUP>

5*(x + 2)<SUP>-3</SUP> = 1

(x + 2)<SUP>-3</SUP> = 1/5

can you finish it now?
 
One note: we do not know that the base is 10.
 
pka said:
One note: we do not know that the base is 10.
Click to expand...

perhaps not explicitly...

but, I always was taught that log a = b meant that a = 10<SUP>b</SUP> And ln a = b meant a = e<SUP>b</SUP> And if the base was something other than 10 or e, it would be specified as in

log<SUB>b</SUB> a = c would mean a = b<SUP>c</SUP>

Maybe I'm assuming too much here.
 
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