How do you solve the equation #log_6(a^2+2)+log_6 2=2#?

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Answer 1 · 2016-10-19T17:02:20

Use the rule #log_a(n) + log_a(m) = log_a(n xx m)# to start the simplification process.

#log_6(2(a^2 + 2)) = 2#

Convert into exponential form using the rule #log_a(n) = b -> a^b = n#

#2a^2 + 4 = 6^2#

#2a^2 + 4 = 36#

#2a^2 - 32 = 0#

#2(a^2 - 16) = 0#

#(a + 4)(a - 4) = 0#

#a = 4 and -4#

Hopefully this helps!

Answer 2 · 2016-10-19T17:08:46

#a=sqrt16=+-4#

#log_6(a^2+2)+log_6(2)=2#

using #color(red)(log_n(X)+log_n(Y)=log_n(XY))#

#log_6(2(a^2+2))=2#

using the definition of logarithms

#color(red)(log_x(y)=z=>x^z=y#

we have

#6^2=2(a^2+2)#

solving:

#2a^2+4=36#

#2a^2=32#

#a^2=16#

#a=sqrt16=+-4#