Logarithmic Equations

arl2267

New member
log4x-log4(x+3)=-1

This is what I have so far:

=log4(x)(x-3)=-1
=(x)(x-3)= 4-1
=(x)(x-3)= 1/4
= x2+3x-1/4=0

Now do I use the quadratic equation to solve? Thanks.

MarkFL

Staff member
We are given to solve:

$\displaystyle \log_4(x)-\log_4(x+3)=-1$

We want to first apply the logarithmic property:

$\displaystyle \log_a(b)-\log_a(c)=\log_a\left(\frac{b}{c} \right)$

What does this give us?

arl2267

New member
I already solved part of the equation, I'm just not sure what to do once I get to

= x2+3x-1/4=0

When I solve using the quadratic equation after this I end up with:

-3 +/- 2squareroot2/2

Jameson

Staff member
I already solved part of the equation, I'm just not sure what to do once I get to

= x2+3x-1/4=0

When I solve using the quadratic equation after this I end up with:

-3 +/- 2squareroot2/2
That's not the equation you should solve.

$\displaystyle \log_4(x)-\log_4(x+3)=-1$. Using MarkFL's suggestion we simplify this to $$\displaystyle \log_{4}\left( \frac{x}{x+3} \right)=-1$$. Using the definition of a logarithm this becomes $$\displaystyle 4^{-1}=\frac{x}{x+3}$$.

Can you finish from here?

arl2267

New member
Do I multiply each side by 1/4?

Jameson

Staff member
$$\displaystyle \frac{1}{4}=\frac{x}{x+3}$$

Here you can simplify a few ways. Maybe you're familiar with the idea of cross-multiplication of fractions.

$$\displaystyle 1(x+3)=4x$$ or simply $$\displaystyle x+3=4x$$