logx+log(x+6=)-log(5x+12)
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[0010001111]... Hello User... [1001010101]
Here's your answer...
A property in logarithms is this...
log x + log y = log xy
We will use this in the sum above.
log x + log (x+6) = log (5x+12)
log (x²+6x) = log (5x+12)
Taking antilog on both sides...
x²+6x = 5x+12
x²+x-12 = 0
(x+4)(x-3) = 0
x = -4 or 3
But log is possible only when x>0
So x≠ -4
x = 3
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Here's your answer...
A property in logarithms is this...
log x + log y = log xy
We will use this in the sum above.
log x + log (x+6) = log (5x+12)
log (x²+6x) = log (5x+12)
Taking antilog on both sides...
x²+6x = 5x+12
x²+x-12 = 0
(x+4)(x-3) = 0
x = -4 or 3
But log is possible only when x>0
So x≠ -4
x = 3
[0110100101]... More questions detected... [010110011110]
//Bot UnknownDude is moving on to more queries
//This is your friendly neighbourhood UnknownDude
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