Question

For what value of a does (one-ninth) Superscript a + 1 Baseline = 81 Superscript a + 1 Baseline times 27 Superscript 2 minus a?

Answer 1

Answer:

$a = -4$

Step-by-step explanation:

The equation to solve for "a" is:

$(\frac{1}{9})^{a+1}=81^{a+1}*27^{2-a}$

The first step is to convert all of them to same bases.

9, 81, and 27, all can be expressed in base 3, lets do this:

$(\frac{1}{9})^{a+1}=81^{a+1}*27^{2-a}\\(3^{-2})^{a+1}=(3^4)^{a+1}*(3^3)^{2-a}$

We can use the property: $(a^b)^c=a^{bc}$ to simplify further:

$(3^{-2})^{a+1}=(3^4)^{a+1}*(3^3)^{2-a}\\3^{-2a-2}=3^{4a+4}*3^{6-3a}$

The right side has 2 same bases multiplied, we can simplify this using the property:  $a^x * a^y = a^{x+y}$

Thus, we have:

$3^{-2a-2}=3^{4a+4}*3^{6-3a}\\3^{-2a-2}=3^{4a+4+6-3a}}\\3^{-2a-2}=3^{a+10}$

Now, both sides have same base, so exponents would be equal. Now lets equate and solve for "a":

$3^{-2a-2}=3^{a+10}\\-2a-2=a+10\\3a=-12\\a=-4$

So,

a = -4

Answer 2

Answer:

-11/3

Step-by-step explanation: