Question

If ${\log }_2(3^{2x-2}+7)=2+{\log }_2(3^{x-1}+1)$ then $x=?$

• A. $4$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

$lo{g}_2(3^{2x-2}+7)=2+lo{g}_2(3^{x-1}+1)$

Since $lo{g}_{c}a+lo{g}_{c}b=lo{g}_c\left(ab\right)$

$lo{g}_2(3^{2x-2}+7)=lo{g}_{2}4+lo{g}_2(3^{x-1}+1)$

$lo{g}_2(3^{2x-2}+7)=lo{g}_2\left(4.\right(3^{x-1}+1\left)\right)$

$3^{(}2x-2)+7=4(3^{x-1}+1\left)\right)$

$3^{(}2x-2)+7=4(3^{x-1})+4$

$3^{(x-1{)}^2}+7-4={4.3}^{x-1}$

Let $3^{x-1}=t$

$t^2+3=4t$

$t^2-4t+3=0$

$(t-1)(t-3)=0$

$t=1;t=3$

$3^{x-1}=1=3^0$ or $3^{x-1}=3$

$x-1=0$ or $x-1=1$

$x=1orx=2$