Solving Algebraic Equations Involving Logarithms

[Corkery]

Homework Statement

Solve Algebraically:

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

Homework 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

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 I am doing it completely wrong. Please help.

 

[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.

 

[VietDao29]

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

(1) log_ab + log_ac = log_a(bc)
(2) log_ab - log_ac = log_a(b / c)
(3) log_a(b^c) = c log_a(b)
(4) The change base formula:
[tex]\log_{a}b = \frac{\log_{c}b}{\log_{c}a}[/tex]

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

You cannot divide log_ab by log_a to get b, it's as meaningless as saying that:
[tex]\frac{\sin x}{\sin} = x[/tex]

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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
[tex]\Leftrightarrow a = 4 \quad \mbox{or} \quad a = -2[/tex]
a² + 2, and 2a + 10 should all be positive. And both values satisfy the requirements, so we have 2 solutions:
[tex]a = 4 \quad \mbox{or} \quad a = -2[/tex]

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

 

[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.

 

[VietDao29]

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.

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