Question

Solution set of the inequality, ${49}^x+7^{x+1}-98<0$, is:

• A. $(-\infty ,\frac{1}{2})$
• B. $(-1,\infty )$
• C. $(-\infty ,1)$
• D. $(-1,\frac{1}{2})$

Answer 1

The correct option is C $(-\infty ,1)$

${49}^x+7^{x+1}-98<0$

$7^{2x}+{7.7}^x-98<0$

$7^{2x}+{14.7}^x-{7.7}^x-98<0$

$(7^x+14)(7^x-7)<0$

$\Rightarrow 7^x<7$ and $7^x<-14$

But $7^x$ is always positive

so $7^x<7$

$x\log 7<\log 7$

i.e, $x<1$

$x\in (-\infty ,1)$