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

[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.