Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1901
2
avatar

log5(7)+log5(3)

 Oct 12, 2014

Best Answer 

 #2
avatar+118696 
+5

log5(7)+log5(3)=log521=log1021log105=1.8917$correctto4decimalplaces$

.
 Oct 12, 2014
 #1
avatar+23254 
+5

Logs are exponents.

You add exponents when you have a multiplication problem with the same bases.

So, if you have added logs, it was a multiplication problem.

log7 + log3  =  log21   (for every base)

Do you need a decimal answer?  If you do, use the change-of-base formula. Any question? Post back.

 Oct 12, 2014
 #2
avatar+118696 
+5
Best Answer

log5(7)+log5(3)=log521=log1021log105=1.8917$correctto4decimalplaces$

Melody Oct 12, 2014

1 Online Users