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:

$2a^2+4=36$

$2a^2=32$

$a^2=16$

$a=sqrt16=+-4$

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