How do you solve #log_(6)x + log_(6)3 = 2#?

1 Answer
Jun 11, 2018

#x=12#

Explanation:

We can start by subtracting #log_6(3)# from both sides. This gives us

#log_6x=color(darkviolet)(2-log_6(3))#

We can apply the logarithm rule

#a=log_b(b^a)#

where our #a# is what I have in purple above. This can be rewritten as

#log_6x=color(darkviolet)(log_6(ul(6^(2-log_6(3))))#

Let's simplify what I have underlined:

#6^(2-log_6(3))# can be rewritten as

#6^(-log_6(3))*6^2#

Which simplifies to

#3^-1*color(blue)(6^2)#

Further simplifying, we get

#1/3*color(blue)(36)=36/3=color(darkviolet)(12)#

Remember, this is the underlined portion of our equation. We have

#log_6x=color(darkviolet)(log_6(12))#

Since the bases are the same, we now have

#x=12#

All we did was:

  • Subtract to get logs on both sides
  • Rewrite the expression in a form we can deal with
  • Simplify the expression using many exponent properties
  • Realized that the bases are the same, so they basically cancel

Hope this helps!