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Guys please help me i cant solve this logarithmic equation

Guys please help me i cant solve this logarithmic equation-example-1

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Answer:

Explanation:

I'll just write log for log base 10. The important rules about logarithms we need to know are:


a \log b = \log( b^a)


\log a + \log b = \log( ab)

and at the end we'll need something like


\log a = \log b \implies a=b

Suitably armed we can begin,


\left( 1 + (1)/(2x) \right) \log 3 + \log 2 = \log (27 - 3^(1/x))

First rule,


\log 3^( 1 + 1/2x)  + \log 2 = \log (27 - 3^(1/x))

Second rule,


\log 2(3^( 1 + 1/2x))  = \log (27 - 3^(1/x))

Third rule,


2(3^( 1 + 1/2x))  = 27 - 3^(1/x)

We can rewrite this to be a quadratic equation in
y=3^(1/2x).


6 (3^(1/2x))  = 27 -  (3^(1/2x))^2


y^2 +6y - 27 =0


(y - 3)(y+9)=0


y=3 \textrm{ or } y=-9

First solution:


3^(1/2x) = 3 = 3^(1)


1/2x = 1


1=2x


x = 1/2

Second solution,


3^(1/2x) = -9

No real powers of 3 will be negative, no solution here.

Answer: x=1/2

Check:

1 + 1/2x = 2 so the left side is log 2(3²) = log 18

Right side, log(27-3^2) = log 18 √

User Electricsheep
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