Logarithm state the value of x for which the equation is defined

[Jaco Viljoen]

Homework Statement

Consider the equation:
3log_x5+2log_x2-log_1/x2=3
a)State which values of x for which the equation is defined.
b)Solve the equation for x.

Homework Equations

The Attempt at a Solution

3log_x5+2log_x2-log_1/x2=3
=log_x5³+log_x2²-log_1/x2=3
=log_x125+log_x4-log_1/x2=3
=log_x500-log_1/x2=3

=log_x2=log_x2/log_x(1/x)change of base
=-log_x2

=log_x125+log_x4+log_x2=3
=log_x500+log_x2=3
=log_x1000=3
=x³=1000
x=10

Please check if you don't mind?

 

[SammyS]

Homework Statement

Consider the equation:
3log_x5+2log_x2-log_1/x2=3
a)State which values of x for which the equation is defined.
b)Solve the equation for x.

Homework Equations

The Attempt at a Solution

3log_x5+2log_x2-log_1/x2=3
=log_x5³+log_x2²-log_1/x2=3
=log_x125+log₄-log_1/x2=3
=log_x500-log_1/x2=3

=log_x2=log_x2/log_x(1/x)change of base
=-log_x2

=log_x125+log₄+log_x2=3
=log_x500+log_x2=3
=log_x1000=3
=x³=1000
x=10

Please check if you don't mind?

Not too difficult to check.

3log_x5+2log_x2-log_1/x2=3

becomes:

3log₁₀5+2log₁₀2-log_1/102=3 .​

See if the following is true.

$\displaystyle\ 10^{\displaystyle\left(3\log_{10}(5)+2\log_{10}(2)-\log_{1/10}(2)\right)}=10^3\$​

 

[Jaco Viljoen]

Yes it works.
I don't understand what 9.1 wants me to do?

 

[SammyS]

Yes it works.
I don't understand what 9.1 wants me to do?

What's 9.1 ?

 

[Jaco Viljoen]

State which values of x for which the equation is defined.

 

[Mark44]

3log_x5+2log_x2-log_1/x2=3
a)State which values of x for which the equation is defined.

For what values of x does log_x(something) make sense? Same question for log_1/x(something).

 

[SammyS]

State which values of x for which the equation is defined.

Several ways to figure this out.

1. From definition of the logarithm.

What does it mean, particularly for base, b, if log_b(A) = C ?​

2. From change of base, along with knowing the domain of the logarithm function.

What are log_x(A) and log_1/x(A) ?​

...

 

[Jaco Viljoen]

log_b(A) = C
b^C=A
x>1

 

[SammyS]

log_b(A) = C
b^C=A
x>1

Doesn't that mean that 1/x < 1 ?

 

[Jaco Viljoen]

x>=3

 

[BvU]

Still let's 1/x < 1 ! But: is that a problem ?

Try to put SammyS' question 1 in words.

 

[SammyS]

log_b(A) = C
b^C=A

What are the restrictions on the base, b, if C and A are to be real numbers?

I think the change of base route might lead to the answer more quickly, but if you're to understand logarithmic & exponential functions, then eventually you need to confront this issue regarding the base.

 

[Jaco Viljoen]

a logarithm is undefined when x<0
1/x will be 1/3 but still x>0 so the logarithm will be defined.
x>0

 

[SammyS]

a logarithm is undefined when x<0
1/x will be 1/3 but still x>0 so the logarithm will be defined.
x>0

I don't see what 1/3 has to do with anything here.

Yes, as you state, "a logarithm is undefined when x<0." ... and, yes, "x > 0" .

Now, are you referring to the base of the logarithm ?

 

[Jaco Viljoen]

Yes, the base of log cannot be smaller than 0

 

[SammyS]

Yes, the base of log cannot be smaller than 0

Can it be zero ?

Why or why not?

 

[Jaco Viljoen]

It is not defined, cannot be calculated, gives an error.
Its a rule

 

[SammyS]

It is not defined, cannot be calculated, gives an error.
Its a rule

What is not defined?

 

[Jaco Viljoen]

the logarithm

 

[SammyS]

the logarithm

Please make a complete statement.

What logarithm is not defined?

 

[Jaco Viljoen]

The real logarithmic function log_b(x) is defined only for x>0. So the base b logarithm of zero is not defined.

 

[SammyS]

The real logarithmic function log_b(x) is defined only for x>0. So the base b logarithm of zero is not defined.

This is precisely why we needed you to make a full statement here.

The question (in Post #1) refers to x.

In this problem, x appears, not as the argument of the logarithm, but appears as the base, as well as 1/x appears as the base.

 

[Jaco Viljoen]

Thank you so much for all your help.
Have a great night.
Jaco

 

[Mark44]

The real logarithmic function log_b(x) is defined only for x>0. So the base b logarithm of zero is not defined.

Is the log function log_b(x) defined if b = 1?

Edit: Fixed my bonehead typo of "if x = 1"

 

[Tony Mondi]

i have tried to follow the steps i am also having a problem in stating the values of x for which the equation is defined

 

[Tony Mondi]

Please Help

 

[SammyS]

i have tried to follow the steps i am also having a problem in stating the values of x for which the equation is defined

Hello Mondi, Welcome to PF.

The original question had to do with the values of x for which $\ \log_x(5)\$ and $\ \log_{1/x}(2)\$ are defined.

Use the change of base formula to help answer this.

 

[Chestermiller]

If I were doing this problem, I would start out by writing
$$3\log_x5=y$$
or, equivalently,
$$\log_x125=y$$
So,
$$x^y=125$$
If I take the natural log of both sides, I get:
$$y=\frac{\ln125}{\ln x}$$
So,
$$3\log_x5=\frac{\ln125}{\ln x}$$
Similarly $$2\log_x2=\frac{\ln4}{\ln x}$$
and
$$\log_{1/x}2=\frac{\ln2}{\ln (1/x)}=-\frac{\ln2}{\ln x}$$

Chet

 

[Jaco Viljoen]

Is the log function log_b(x) defined if x = 1?

Edit (Mark44): I meant "if b = 1". I have since fixed what I wrote earlier. The base can't be zero and it can't be 1.
Hi Mark,
I thought so but when I checked, NO.
The rule states:
x>0 and not = to 1

Thank you for pointing that out.
Jaco

 

[HallsofIvy]

Hi Mark,
I thought so but when I checked, NO.
The rule states:
x>0 and not = to 1

Thank you for pointing that out.
Jaco

This is the second time you have posted this and it still is NOT true. Where did you find that rule? For any base, b, b^0= 1 so log_b(1)= 0. That certainly is defined!

 

[SammyS]

Hi Mark,
I thought so but when I checked, NO.
The rule states:
x>0 and not = to 1

Thank you for pointing that out.
Jaco

The base, b, must be positive, but can't be 1. ( I suppose Mark had a typo.)

 

[Jaco Viljoen]

This is the second time you have posted this and it still is NOT true. Where did you find that rule? For any base, b, b^0= 1 so log_b(1)= 0. That certainly is defined!

Hi Hallsoflvy,
I am not referring to log_bx but to the problem where x is the base:
3log_x5+2log_x2-log_1/x2=3

Sammy,
I understood what Mark was saying the base must be positive but can't be 1.

 

[Chestermiller]

I see that you liked what I posted in #28. Does that mean that you figured out how to use it to solve the problem? If so, please show us what you did.

Chet

 

[Jaco Viljoen]

I see that you liked what I posted in #28. Does that mean that you figured out how to use it to solve the problem? If so, please show us what you did.

Chet

Hi Chet,
I was thanking you for your interest and alternative option, I will however give it a bash:
what rule did you apply or why did you use ln?

$$\frac{\ln125}{\ln x}+\frac{\ln4}{\ln x}+\frac{\ln2}{\ln x}=3$$
$$\frac{\ln1000}{\ln x}=3$$
$$ln1000=3lnx$$
ln(x)³=ln1000
x³=1000
10³=1000
so x=10I am not sure if I have done the right thing... Please advise

 

[Chestermiller]

Hi Chet,
I was thanking you for your interest and alternative option, I will however give it a bash:
what rule did you apply or why did you use ln?

$$\frac{\ln125}{\ln x}+\frac{\ln4}{\ln x}+\frac{\ln2}{\ln x}=3$$
$$\frac{125}{\ x}+\frac{4}{\ x}+\frac{2}{\ x}=3$$
$$\frac{125+4+2}{\ x}=3$$
$$\frac{131}{\ x}=3$$
$$131=3x$$
$$\frac{131}{\ 3}=x$$
$$x=43\frac{2}{\ 3}$$

I am not sure if I have done the right thing... Please advise

Well, your first equation is correct, but that's about all. Going from your first equation to your second equation is definitely not how logs work.

Did you notice that the three terms on the left hand side of your first equation have a common denominator. What do you usually do in an algebra problem or arithmetic problem when you have the sum of three separate terms, all of which have the same denominator?

Chet

 

[Raghav Gupta]

$$\frac{\ln125}{\ln x}+\frac{\ln4}{\ln x}+\frac{\ln2}{\ln x}=3$$
$$\frac{\ln1000}{\ln x}=3$$I am not sure if I have done the right thing... Please advise

Yeah, this is correct.
Do, some further solving.

 

[Jaco Viljoen]

$$\frac{\ln125}{\ln x}+\frac{\ln4}{\ln x}+\frac{\ln2}{\ln x}=3$$
$$\frac{\ln1000}{\ln x}=3$$
$$ln1000=3lnx$$
ln(x)³=ln1000
x³=1000
10³=1000
so x=10

Boom, done.

I don't think this change of base rule was covered in my manual.
Can this always be used?

Thank you

 

[Mark44]

The base, b, must be positive, but can't be 1. ( I suppose Mark had a typo.)

That's what I meant, but was somehow unable to make my fingers follow the instructions from my brain. Thanks for pointing it out, Sammy! Embarassing, but I would rather see a mistake be corrected.

In my defense, the original equation had x as the base:

3log_x5+2log_x2-log_1/x2=3

 

[Raghav Gupta]

I don't think this change of base rule was covered in my manual.
Can this always be used?

Thank you

You could use that anytime.Of whether it is useful or not depends on the situation and your thinking.