Question

Number of real values of $x$ satisfying $x+{\log }_{10}(1+2^x)=x{\log }_{10}5+{\log }_{10}6$ is

• A. $0$
• B. $1$
• C. $2$
• D. More than $2$

Answer 1

Answer: (B)

$1$

$x+\log (1+2^x)=x\log 5+\log 6⟹x-x\log 5=\log 6-\log (1+2^x)⟹x(1-\log 5)=\log 6-\log (1+2^x)⟹x(\log 10-\log 5)=\log 6-\log (1+2^x)⟹x\log \left(\frac{10}{5}\right)=\log \left(\frac{6}{1+2^x}\right)⟹x\log 2=\log \left(\frac{6}{1+2^x}\right)⟹\log 2^x=\log \left(\frac{6}{1+2^x}\right)⟹2^x=\frac{6}{1+2^x}⟹2^x(1+2^x)=6⟹2^x+{\left(2^x\right)}^2=6⟹2^x+2^{2x}=6⟹2^x+2^{2x}-6=0$

Take $m=2^x$, then we have:

$⟹m^2+m-6=0⟹m^2+3m-2m-6=0⟹m(m+3)-2(m+3)=0⟹(m-2)(m+3)=0⟹m-2=0andm+3=0⟹m=2andm=-3\therefore 2^x=2and{2}^x=-3$

$⟹2^x=2^{1}and{2}^x=-3$ is invalid.

$⟹x=1$

$\therefore$ No. of real values of x = 1.