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sauravshakya
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HINT: logxlog6 = log(x/6)

sauravshakya
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Now, can u?

TheViper
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so \(\frac{x}{6}=15\) ??
@sauravshakya

sauravshakya
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Yep... @TheViper

TheViper
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Yeah I hoped :)
THANX ;)

sauravshakya
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WElcome

vishweshshrimali5
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@yeller33
see,
log a log b = log(a/b)
and
log a + log b = log (ab)
remember these 2 formula for ever

abayomi12
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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


sauravshakya
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dw:1345303238963:dwAnd , right? @vishweshshrimali5

vishweshshrimali5
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Yes @sauravshakya absolutely correct

yeller33
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okay i understand thanks guys!

sauravshakya
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Welcome

yeller33
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how do you do it if it has a number at the end. = log7+logx=2.... the 2 throws me off.

Kaederfds
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x is 90

sauravshakya
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log7 + logx=2
log(7*x)=2

sauravshakya
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Does it help?

vishweshshrimali5
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and remember that
\[\large log_a a = 1\]

yeller33
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x=2//7??
( 7x=2 )

vishweshshrimali5
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Thus u can write
2 = 2* 1 = \(\large 2* log_{10} 10\)
also

sauravshakya
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log7 + logx=2
log(7*x)=2
10^2 = 7*x
100/7=x

vishweshshrimali5
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\[\large b\space log _a x = log_a (b^x)\]

vishweshshrimali5
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That's how @sauravshakya got the third step

sauravshakya
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got it? @yeller33

vishweshshrimali5
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i can write
2 = \(\large \log_{10} (10^2)\)

sauravshakya
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Yep @vishweshshrimali5


sauravshakya
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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

sauravshakya
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according to @vishweshshrimali5

yeller33
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can you two see if i did this one correctly:
logx+log8=1
log8*logx=1
8x=1
8x=10^1
10/8

yeller33
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i tried to follow your explanations.

sauravshakya
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dont think so

sauravshakya
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your second step is wrong

sauravshakya
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logx+log8= log8x

sauravshakya
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and 1=log10

sauravshakya
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got your mistake?

yeller33
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i think so . thanks again!