Logarithm state the value of x for which the equation 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.