Fun with logs

Algebra Level pending

$2log_5{(2^x-5)}=log_5{2}+log_5\left(2^x-\frac{7}{2}\right)$ Evaluate $x$ .

6 1 5 3;2 3 0

1 solution

Gian Sanjaya
Sep 5, 2015

Assume $2^x=y$ . Then, by applying 5^LHS = 5^RHS and simplifying it:

$(y-5)^2=2(y-\frac{7}{2})$

$y^2-12y+32=0$

$(y-4)(y-8)=0$

However, we need $y-5>0$ ,hence $y>5$ , so $y=8$ . This results $x=\boxed{3}$ .

clean solution $\ddot\smile$ !

Rohit Udaiwal - 5 years, 9 months ago