Answer:
Step-by-step explanation:
I'll just write log for log base 10. The important rules about logarithms we need to know are:
[tex] a \log b = \log( b^a) [/tex]
[tex] \log a + \log b = \log( ab) [/tex]
and at the end we'll need something like
[tex]\log a = \log b \implies a=b[/tex]
Suitably armed we can begin,
[tex] \left( 1 + \dfrac{1}{2x} \right) \log 3 + \log 2 = \log (27 - 3^{1/x}) [/tex]
First rule,
[tex] \log 3^{ 1 + 1/2x} + \log 2 = \log (27 - 3^{1/x}) [/tex]
Second rule,
[tex] \log 2(3^{ 1 + 1/2x}) = \log (27 - 3^{1/x}) [/tex]
Third rule,
[tex] 2(3^{ 1 + 1/2x}) = 27 - 3^{1/x} [/tex]
We can rewrite this to be a quadratic equation in [tex]y=3^{1/2x}[/tex].
[tex] 6 (3^{1/2x}) = 27 - (3^{1/2x})^2 [/tex]
[tex]y^2 +6y - 27 =0[/tex]
[tex](y - 3)(y+9)=0[/tex]
[tex]y=3 \textrm{ or } y=-9[/tex]
First solution:
[tex]3^{1/2x} = 3 = 3^{1}[/tex]
[tex]1/2x = 1[/tex]
[tex]1=2x[/tex]
[tex]x = 1/2[/tex]
Second solution,
[tex]3^{1/2x} = -9[/tex]
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 √
