help please w/ solving (1/4)(10)^(1/x) = (10)^(1/(2x)) + 15

Re: some more help with exponential

Hello, oded244!

Wow ... the problems are getting uglier!


1410x  =  102x+15\displaystyle \frac{1}{4}\sqrt[x]{10} \;=\;\sqrt[2x]{10} + 15

We have:   101x41012x60  =  0\displaystyle \text{We have: }\;10^{\frac{1}{x}} - 4\cdot10^{\frac{1}{2x}} - 60 \;=\;0

. . \(\displaystyle \text{a quadratic! }\hdots\quad \left(10^{\frac{1}{2x}}\right)^2 - 4\left(10^{\frac{1}{2x}}\right) - 60 \;=\;0\)

. . which factors:   (1012x+6)(1012x10)  =  0\displaystyle \text{which factors: }\;\left(10^{\frac{1}{2x}} + 6\right)\,\left(10^{\frac{1}{2x}} - 10\right) \;=\;0


\(\displaystyle 10^{\frac{1}{2x}} + 6 \:=\:0 \quad\Rightarrow\quad 10^{\frac{1}{2x}} \:=\:-6 \quad\Rightarrow\quad \frac{1}{2x} \:=\:\log(-6) \quad\hdots\;\;\text{No real root}\)

1012x10=01012x=10112x=1x=12\displaystyle 10^{\frac{1}{2x}} - 10 \:=\:0 \quad\Rightarrow\quad 10^{\frac{1}{2x}} \:=\:10^1 \quad\Rightarrow\quad \frac{1}{2x} \:=\:1 \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}


Edit: corrected an error
 
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