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Solve log52 (x-3) = x

  • Thread starter Acnhduy
  • Start date
  • #1
31
1

Homework Statement



log52 (x-3) = x

Homework Equations


log laws


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 :)
 

Answers and Replies

  • #2
DrClaude
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You can graph a function, not an equation.
 
  • #3
31
1
OH i see thank you
 
  • #4
haruspex
Science Advisor
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You can graph a function, not an equation.
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
31
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I was given (log52) (x-3)
 
  • #6
575
76
You can graph a function, not an equation.
This is not true. The graph of an equation is the set of tuples which make the equation a true statement.
 
  • #7
31
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How do I graph this?
 
  • #8
575
76
How do I graph this?
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
31
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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 :frown:

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
575
76
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 :frown:

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.
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
31
1
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
##x=\frac{3a}{a-1}=\frac{3\log_5 2}{\log_5 2-1}##.
 
  • #12
575
76
...
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
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
31
1
Okay then, thanks for the help !
 
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