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HINT: logx-log6 = log(x/6)

Now, can u?

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

Yeah I hoped :)
THANX ;)

WElcome

|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

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

Yep @vishweshshrimali5

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!