help please w/ solving (1/4)(10)^(1/x) = (10)^(1/(2x)) + 15
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Aug 18, 2008 #2 Re: some more help with exponential Hello, oded244! Wow ... the problems are getting uglier! \(\displaystyle \frac{1}{4}\sqrt[x]{10} \;=\;\sqrt[2x]{10} + 15\) Click to expand... \(\displaystyle \text{We have: }\;10^{\frac{1}{x}} - 4\cdot10^{\frac{1}{2x}} - 60 \;=\;0\) . . \(\displaystyle \text{a quadratic! }\hdots\quad \left(10^{\frac{1}{2x}}\right)^2 - 4\left(10^{\frac{1}{2x}}\right) - 60 \;=\;0\) . . \(\displaystyle \text{which factors: }\;\left(10^{\frac{1}{2x}} + 6\right)\,\left(10^{\frac{1}{2x}} - 10\right) \;=\;0\) \(\displaystyle 10^{\frac{1}{2x}} + 6 \:=\:0 \quad\Rightarrow\quad 10^{\frac{1}{2x}} \:=\:-6 \quad\Rightarrow\quad \frac{1}{2x} \:=\:\log(-6) \quad\hdots\;\;\text{No real root}\) \(\displaystyle 10^{\frac{1}{2x}} - 10 \:=\:0 \quad\Rightarrow\quad 10^{\frac{1}{2x}} \:=\:10^1 \quad\Rightarrow\quad \frac{1}{2x} \:=\:1 \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}\) Edit: corrected an error
Re: some more help with exponential Hello, oded244! Wow ... the problems are getting uglier! \(\displaystyle \frac{1}{4}\sqrt[x]{10} \;=\;\sqrt[2x]{10} + 15\) Click to expand... \(\displaystyle \text{We have: }\;10^{\frac{1}{x}} - 4\cdot10^{\frac{1}{2x}} - 60 \;=\;0\) . . \(\displaystyle \text{a quadratic! }\hdots\quad \left(10^{\frac{1}{2x}}\right)^2 - 4\left(10^{\frac{1}{2x}}\right) - 60 \;=\;0\) . . \(\displaystyle \text{which factors: }\;\left(10^{\frac{1}{2x}} + 6\right)\,\left(10^{\frac{1}{2x}} - 10\right) \;=\;0\) \(\displaystyle 10^{\frac{1}{2x}} + 6 \:=\:0 \quad\Rightarrow\quad 10^{\frac{1}{2x}} \:=\:-6 \quad\Rightarrow\quad \frac{1}{2x} \:=\:\log(-6) \quad\hdots\;\;\text{No real root}\) \(\displaystyle 10^{\frac{1}{2x}} - 10 \:=\:0 \quad\Rightarrow\quad 10^{\frac{1}{2x}} \:=\:10^1 \quad\Rightarrow\quad \frac{1}{2x} \:=\:1 \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}\) Edit: corrected an error
