How do you solve for x in log_2x-log_2 3=5log2xlog23=5?

1 Answer

x=96x=96

Explanation:

Let's first combine the left side, using the rule that log_x(a)-log_x(b)=log_x(a/b)logx(a)logx(b)=logx(ab):

log_2(x)-log_2(3)=5log2(x)log2(3)=5

log_2(x/3)=5log2(x3)=5

We can now set these as exponents to a base of 2 (to eliminate the log):

2^(log_2(x/3))=2^52log2(x3)=25

On the left side, this has the effect of taking the inverse function of the log and so cancels out, leaving:

x/3=32x3=32

x=96x=96

And now let's check it:

log_2(96)-log_2(3)=5=>6.585-1.585=5log2(96)log2(3)=56.5851.585=5

(the logs are rounded to 3 decimal places)