How to Solve x: log10(10x/3) - 2 = log10(1/10)?

1 Answer
Feb 26, 2018

#x=3#

Explanation:

I am taking these to be base 10 logs:

#log(a/b)=loga-logb#

#log_10((10x)/3)-2=log_(10)(1/10)-log#

#log_10((10x)/3)-2=log_(10)(1)-log_(10)(10)#

#log_10((10x)/3)-2=0-1#

Note:

#log_(10)(10)=1# (the logarithm of the base is always 1)

#log_(10)(1)=0# (the logarithm of 1 is always 0, whatever the base)

#log_10((10x)/3)-2=0-1#

#log_10((10x)/3)-2=-1#

#log_10((10x)/3)=1#

Using these as powers of #10#

#10^(log_10((10x)/3))=10^1#*

#(10x)/3=10#

#x=30/10=3#

*
If:

#y=log_b(a)#

Then:

#b^y=a#

#:.#

#b^(log_b(a))=a#