logx-log6=log15

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

logx-log6=log15

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

HINT: logx-log6 = log(x/6)
Now, can u?
so \(\frac{x}{6}=15\) ?? @sauravshakya

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Yep... @TheViper
Yeah I hoped :) THANX ;)
WElcome
@yeller33 see, log a- log b = log(a/b) and log a + log b = log (ab) remember these 2 formula for ever
log(x) - log(6) = log(15) Add log(6) to each side: log(x) = log(15) + log(6) = log(15 times 6) x = 15 times 6 x = 90
Gud
|dw:1345303238963:dw|And , right? @vishweshshrimali5
Yes @sauravshakya absolutely correct
okay i understand thanks guys!
Welcome
how do you do it if it has a number at the end. = log7+logx=2.... the 2 throws me off.
x is 90
log7 + logx=2 log(7*x)=2
Does it help?
and remember that \[\large log_a a = 1\]
x=2//7?? ( 7x=2 )
Thus u can write 2 = 2* 1 = \(\large 2* log_{10} 10\) also
log7 + logx=2 log(7*x)=2 10^2 = 7*x 100/7=x
\[\large b\space log _a x = log_a (b^x)\]
That's how @sauravshakya got the third step
got it? @yeller33
i can write 2 = \(\large \log_{10} (10^2)\)
:)
log7 + logx=2 log(7*x)=2 1/2 log(7x)=1 log{(7x)^1/2} = log10 (7x)^(1/2) = 10 7x = 100 x=100/7
according to @vishweshshrimali5
can you two see if i did this one correctly: logx+log8=1 log8*logx=1 8x=1 8x=10^1 10/8
i tried to follow your explanations.
dont think so
your second step is wrong
logx+log8= log8x
and 1=log10
got your mistake?
i think so . thanks again!

Not the answer you are looking for?

Search for more explanations.

Ask your own question