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

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Other answers:

Yep... @TheViper
Yeah I hoped :) THANX ;)
@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
|dw:1345303238963:dw|And , right? @vishweshshrimali5
Yes @sauravshakya absolutely correct
okay i understand thanks guys!
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!

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