Think!

Algebra Level 3

$\large \left( 1+\dfrac{1}{2x} \right) \log 3 +\log 2=\log \left ( 27-\sqrt[x]{3} \right )$

Find the number of value(s) of $x$ satisfying the equation above.

The answer is 1.

2 solutions

Sabhrant Sachan
May 18, 2016

We have $\left(1+\dfrac{1}{2x}\right)\times\log{3} +\log{2} = \log{(27-\sqrt[x]{3})}$

$\log{3}+\dfrac{1}{2x}\times\log{3} +\log{2} = \log{(27-\sqrt[x]{3})} \\ \log{6}+\log{\sqrt[2x]{3}} = \log{(27-\sqrt[x]{3})}$

Put $\sqrt[2x]{3} = a$ , so $\sqrt[x]{3}=a^2$

$\log{6a}=\log{(27-a^2)} \\ a^2+6a-27=0 \\ a^2+9a-3a-27=0 \\ (a+9)(a-3)=0 \\ a=-9,3 ,\text{but } a\ne -9 \text{ as } a>0 \\ \implies x=\dfrac12 \text{ but } x \in N \text{ for } \sqrt[x]{3} \text { to be defined }$

Hence , The equation has no Solutions

Moderator note:

(May 30, 2021)

because $x^y = \sqrt[1/y]{x}$

$\sqrt[1/2]{3} = 3^{1/(1/2)} = 3^2 = 9$

So $x = \frac12$ is a solution.

Well $x=1/2$ satisfy the given eqn.. So I guess you should make some edits in your solution and mention regarding the eqn having one solution.

Rishabh Jain - 5 years ago

$\sqrt[x]{3} \Rightarrow x\in \mathbb{N}$ , hence no solution!

Akshat Sharda - 5 years ago

What??

Rishabh Jain - 5 years ago

@Rishabh Jain – Although $\sqrt[x]{3}=3^{\frac{1}{x}}$ but still in our first expression we are given $\sqrt[x]{3}\Rightarrow \huge x\in \mathbb{N}$ .

If the question had $\left( 1+\dfrac{1}{2x} \right) \log 3 +\log 2=\log( 27-3^{\frac{1}{x}})$ then 1 solution would be $x=\frac{1}{2}$ .

Akshat Sharda - 5 years ago

@Akshat Sharda – For $a^x$ , the exponential function for any $a \in \mathbb{C}$ , we don't have any restrictions like $x \in \mathbb{N}$ . If $x$ is of form $\dfrac 1y$ then this simply converts to $\sqrt[y]{a}$ .

If that is true, then why $\sqrt[x]{a}$ and $a^{{1}/{x}}$ are not synonymous to each other regarding the solutions for $x$ .

Tapas Mazumdar - 4 years, 2 months ago

I'm not sure I consider this a math question or a question about understanding notation. When I worked on the problem I converted the radical sign to 1/x and found there would be one solution without finding the solution. I wasn't aware that when using the radical symbol, the index needed to be a natural number. It does make sense. I decided to Google this. It took a long time to find a definition that said the index had to be a natural number. A lot implied it but didn't say it and many didn't mention it at all.

John Morrison - 1 year, 6 months ago

Hana Wehbi
May 18, 2016

( $1+\frac{1}{2x}$ ) $log(3)+log(2)$ = $log(27-(3)^{1/x}$ ). Subtract $log(2)$ from both sides and divide by $log(3)$

( $1+\frac{1}{2x}$ ) = $\frac{log(27-3^{1/x})}{log 3}$ - $\frac{log(2)}{log(3)}$

$\frac{log 2}{log 3}$ = $1.585$ ;

$2.585 =$ $\frac{log(27-3^{1/x})}{log 3}$ - $\frac{1}{2x}$ ;

$2.585 =$ $log_{3}(27-3^{1/x}$ ) - $\frac{1}{2x}$ ;

$2.585 =$ $log_{3}(3^{3}(1-3^{(1/x)-3}$ )) - $\frac{1}{2x}$ ;

$2.585 =$ $3(1-3^{(1/x)-3}$ ) - $\frac{1}{2x}$ ;

From here, we can tell that there is no Natural value that satisfies the equation or there are 0 values in set N.

Moderator note:

(May 30, 2021)

because $x^y = \sqrt[1/y]{x}$

$\sqrt[1/2]{3} = 3^{1/(1/2)} = 3^2 = 9$

So $x = \frac12$ is a solution.

Think!

The answer is 1.

2 solutions

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