OpenStudy (anonymous):

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

5 years ago
OpenStudy (ghazi):

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

5 years ago
OpenStudy (anonymous):

yes, thank you

5 years ago
OpenStudy (ghazi):

YW :)

5 years ago
OpenStudy (anonymous):

Is the answer 13?

5 years ago
OpenStudy (ghazi):

i guess no

5 years ago
OpenStudy (anonymous):

Then I'm confused

5 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

5 years ago
OpenStudy (anonymous):

I'll try

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

How do I plug the problem into that?

5 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?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

yep...

5 years ago
OpenStudy (anonymous):

thanks

5 years ago
OpenStudy (anonymous):

yw...:)

5 years ago
OpenStudy (anonymous):

So its x =15

5 years ago