Question

Guys please help me i cant solve this logarithmic equation

Answer 1

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 √