How to Solve Logarithmic Equation log4x-log4(x+3)=-1?

[arl2267]

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]

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]

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]

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 [math]\log_{4}\left( \frac{x}{x+3} \right)=-1[/math]. Using the definition of a logarithm this becomes [math]4^{-1}=\frac{x}{x+3}[/math].

Can you finish from here?

 

[arl2267]

Do I multiply each side by 1/4?

 

[Jameson]

[math]\frac{1}{4}=\frac{x}{x+3}[/math]

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

[math]1(x+3)=4x[/math] or simply [math]x+3=4x[/math]

 

[arl2267]

So the answer is x=1?

 

[Jameson]

So the answer is x=1?

(Yes)

Wolfram confirms