# Solve log52 (x-3) = x

1. Oct 3, 2013

### Acnhduy

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

log52 (x-3) = x

2. Relevant equations
log laws

3. The attempt at a solution
One of my friends came up with this equation and asked me to graph it. I haven't seen an equation like this, and do not know how to graph using mapping notation or any other methods. Is it possible to graph this?

Looking at the equation, I would assume that there can only be one answer to this because

log52 can only have one answer since this means that 5x = 2

Can someone please point me in the right direction?

Thanks :)

2. Oct 3, 2013

### Staff: Mentor

You can graph a function, not an equation.

3. Oct 3, 2013

### Acnhduy

OH i see thank you

4. Oct 3, 2013

### haruspex

True, but maybe what was meant was to graph the function each side of the equation and see where the graphs cross.
Acnhduy, did you mean (log52) (x-3) or log5(2 (x-3))?

5. Oct 3, 2013

### Acnhduy

I was given (log52) (x-3)

6. Oct 3, 2013

### gopher_p

This is not true. The graph of an equation is the set of tuples which make the equation a true statement.

7. Oct 3, 2013

### Acnhduy

How do I graph this?

8. Oct 3, 2013

### gopher_p

If you really were given the equation $\log_5(2)\cdot(x-3)=x$, then you have a linear equation in a single variable. It has one solution (value which makes it true); $\frac{3\log_5(2)}{\log_5(2)-1}$. The graph is the singleton set $\{\frac{3\log_5(2)}{\log_5(2)-1}\}$. The picture associated with this graph is rather uninteresting; it's just a point plotted on the real line.

9. Oct 3, 2013

### Acnhduy

How did you get to $\frac{3\log_5(2)}{\log_5(2)-1}$ ?

Is it because you distribute log52 to (x-3)?

Can you show me the steps, I don't really know what is going on

If you distribute it, all I can think of is

x = xlog52 - 3log52

the coefficient can become the exponent so

x = log52x - log523

if so, since the bases are the same I can divide?

x= log5 (2x / 8)

Please show me the steps, this is all I can think of.

Many thanks.

10. Oct 3, 2013

### gopher_p

Keep in mind that $\log_5 2$ is just a number. For the time being, let's use the letter $a$ in its place. So you have

$a(x-3)=x$

which after distributing the $a$ becomes

$ax-3a=x$.

Grouping the $x$ terms on the left and everything else on the right gives

$ax-x=3a$,

and factoring out $x$

$(a-1)x=3a$.

Solving for $x$ and plugging $\log_5 2$ back in for $a$ gives

$x=\frac{3a}{a-1}=\frac{3\log_5 2}{\log_5 2-1}$.

11. Oct 3, 2013

### Acnhduy

Oh! I understand now ! :D

Would doing this result in the same thing?

log52 (x-3) = x

Can I divide both sides by (x-3)

log52 = x / (x-3)

And now use the definition of log where

5x/(x-3) = 2

I don't know where I am going with this but how do I solve for x in this case, and would it give me the same as

12. Oct 3, 2013

### gopher_p

Not to be flippant, but in order to solve $5^{\frac{x}{x-3}}=2$, you undo all the things you just did and then do it the way that I showed. There is really no way to do it that involves using any properties of logarithms or exponents.

If this were an actual problem from a precalc class, it's likely that the problem started with $5^{\frac{x}{x-3}}=2$ and asked you to solve for $x$.

13. Oct 3, 2013

### Acnhduy

Okay then, thanks for the help !