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+2 votes
Please help with this question:-

in Numerical Ability by Boss (13.8k points)
edited by | 316 views
no real $x$ is exised for the given equation.

If $x$ is a positive non-zero real number, Now in equation $(\sqrt{243}+3)^x+(\sqrt{243}-3)^x=15^x $, left part is $(15.58...+3)^x +$ some +ve value $= 15^x$, left hand side is greater than right  hand side. So, no +ve real 'x' exist here. when $x=0$, then also equation is not satisfied. when $x$ is negative then value of $15^x$  will always be less than $(\sqrt{243}+3)^x+(\sqrt{243}-3)^x$ and values of both will be tending to zero but value can't be same for any negative $x$.
$\left ( \frac{\sqrt{243}+3}{15} \right )^{x}+\left ( \frac{\sqrt{243}-3}{15} \right )^{x}=1$

Now, putting $x$ anything, we cannot get a $1$ on RHS


@ankitgupta.1729,Ideally if we go to see,in a way since here 'x' hasn't been restricted to Real Numbers or for that matter Natural Numbers you've assumed x for 'Real' values. Isn't it?


@Devshree Dubey Yeah, I have assumed that you were asking solution for real values of x..if there is no restriction then we have to take complex roots as mentioned in the above link which I guess difficult to find.

2 Answers

0 votes
After applying properties of log and doing all calculation I am x=-1.20 negative value and on above equation, if you put x=1 it will give 234/15 which is not equal to 1,


sorry x=0 will not be a solution here I wrongly mentioned it
by (289 points)
edited by
how x=0, can satisfy ?
0 votes

$√243+3 = 12.58/15 <1 , √243+3 = 18.58/15 >1$

$(\frac{√243+3}{15})^x$ will be >1 if X >1

$(\frac{√243-3}{15})^x$ will be <1 always

so sum of these two cannot be 1 for any real positive number.
by Boss (16.8k points)
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