# Log equations

1. Apr 23, 2007

### Corkery

1. The problem statement, all variables and given/known data
Solve Algebraically:

Log(subscript4) (a^2 + 2) = Log(subscript 4) (2a + 10)

2. Relevant equations
other equations that are similar which I can't do are:
Solve Algebraically:

-Log(subscript 5) (x + 3) + Log(subscript 5) (x - 2) = Log(subscript 5) 14

-Log(subscript 2) (2x + 6) - Log(subscript 2) x = 3

3. The attempt at a solution
not sure if this is right but I think you would divide by Log(subscript 4)
Log(subscript4) (a^2 + 2) = Log(subscript 4) (2a + 10)
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Log(subscript4) Log(subscript4)

canceled those out leaving:
a^2 + 2 = 2a + 10

expand the equation:
this is where i get lost, unless im doing it completely wrong. Please help.

2. Apr 23, 2007

### Dick

Your solution is essentially right, but the formalism is painfully wrong. If f(x)=f(y) you cannot divide by f. You have to apply the inverse function of f. The inverse of Log(subscript4) is 4^. As in 4^(Log(subscript4)(x))=x. And a^2 + 2 = 2a + 10
is a harmless quadratic equation. Don't get confused there, just factor it or use the quadratic equation.

3. Apr 23, 2007

### VietDao29

When doing log equations, you should try to remember all the formulae. Some useful are:

(1) logab + logac = loga(bc)
(2) logab - logac = loga(b / c)
(3) loga(bc) = c loga(b)
(4) The change base formula:
$$\log_{a}b = \frac{\log_{c}b}{\log_{c}a}$$

(5) Since logarithm is either strictly increasing or strictly decreasing (i.e, 1-1 function), we have:
$$\log_a b = \log_a c \Leftrightarrow b = c$$

You cannot divide logab by loga to get b, it's as meaningless as saying that:
$$\frac{\sin x}{\sin} = x$$

----------------

I'll help you do the first one, and let you try the other 2 by yourself. See if you can get it. :)

log4 (a2 + 2) = log4 (2a + 10)
Using (5), we have:
<=> a2 + 2 = 2a + 10
<=> a2 - 2a - 8 = 0
$$\Leftrightarrow a = 4 \quad \mbox{or} \quad a = -2$$
a2 + 2, and 2a + 10 should all be positive. And both values satisfy the requirements, so we have 2 solutions:
$$a = 4 \quad \mbox{or} \quad a = -2$$

Is it cear?
Can you do the other 2? :)

Last edited: Apr 23, 2007
4. Apr 23, 2007

### AKG

VietDao, why should a be positive? You need a2+2 and 2a+10 to be positive, you don't need a to be positive. Note, those expressions are positive for both a=4 and a=-2.

5. Apr 23, 2007

### VietDao29

Whoops, I thought that a was the base . Er... stand corrected.
Thanks.