Solve for x?

If #y = a log_10 x#, find x when a = 2 and y = 3

2 Answers
Jun 5, 2018

I tried this:

Explanation:

Let us rearrange it and use our values:

#log_(10)x=y/a#

#log_(10)x=3/2#

use the definition of log:

#x=10^(3/2)#

use the property of exponents:

#x=sqrt(10^3)#

giving:

#x=10sqrt(10)=31.62#

Jun 5, 2018

#x approx 31.622777#

Explanation:

#3 = 2log_10 x#

#log_10 x= 3/2#

#x= 10 ^(3/2)#

#x approx 31.622777#