How do you solve #(\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}#?

Answer 1

#x=-4# or #x=-2#

Work on the left side first:

Since #81=9^2# we know #1/81=1/9^2# and by the rules #1/a=a^-1#, we know that #1/9^2 = 9^-2#
By the rule #(a^b)^c=a^(bc)#, we know
#(9^-2)^(6x+2) =9^(-12x-4)#.

Our equation now appears as follows:

#9^(-12x-4)=9^(2x^2+12)#
If #a^b = a^c#, then #b=c#, so:
#-12x-4 = 2x^2+12#

putting everything aside and gathering in the manner described by our terms:

#2x^2+12x+16=0#

Splitting the whole by two:

#x^2+6x+8=0#

factoring

#(x+4)(x+2)=0#

Solving:

#x=-4# or #x=-2#
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Answer 2

To solve the equation ((\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}), take the logarithm of both sides with the same base, preferably the natural logarithm. Then solve for (x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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