Mathematics
OpenStudy (anonymous):

which of the following is the solution of $\log _{x-5}0.001= -3$

6 years ago
OpenStudy (ghazi):

you have to use base change formula... Log(0.0001)/log(x-5)=-3....can you do it now?

6 years ago
OpenStudy (anonymous):

yes, thank you

6 years ago
OpenStudy (ghazi):

YW :)

6 years ago
OpenStudy (anonymous):

6 years ago
OpenStudy (ghazi):

i guess no

6 years ago
OpenStudy (anonymous):

Then I'm confused

6 years ago
OpenStudy (ghazi):

just you can equate log both the sides by $\log(10^{-4})= \log(x-5)^3....(x-5)^3=0$ now i hope you can do this

6 years ago
OpenStudy (anonymous):

I'll try

6 years ago
OpenStudy (anonymous):

That's a bit too complicated. Using the properties of log, $log_ba=c$$a=b^c$$\sqrt[c]{a}=b$

6 years ago
OpenStudy (anonymous):

How do I plug the problem into that?

6 years ago
OpenStudy (anonymous):

$$\huge log_{x-5} (0.001)=-3$$ means $$\huge (x-5)^{-3}=0.001$$ so can you solve for x from here?

6 years ago
OpenStudy (anonymous):

Do I multiply x-5 to the negative 3rd power? or do I solve using "Solving Equation with Unequal bases?

6 years ago
OpenStudy (anonymous):

nah.... just change that right side to a power of 10... $$\huge (x-5)^{-3}=0.001$$ $$\huge (x-5)^{-3}=10^{-3}$$ $$\huge x-5=10$$

6 years ago
OpenStudy (anonymous):

after that do I add 5 to each side so it will become$x=15$

6 years ago
OpenStudy (anonymous):

yep...

6 years ago
OpenStudy (anonymous):

thanks

6 years ago
OpenStudy (anonymous):

yw...:)

6 years ago
OpenStudy (anonymous):

So its x =15

6 years ago