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

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

6 years agoyes, thank you

6 years agoYW :)

6 years agoIs the answer 13?

6 years agoi guess no

6 years agoThen I'm confused

6 years agojust 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 agoI'll try

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

6 years agoHow do I plug the problem into that?

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

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

6 years agonah.... 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 agoafter that do I add 5 to each side so it will become\[x=15\]

6 years agoyep...

6 years agothanks

6 years agoyw...:)

6 years agoSo its x =15

6 years ago