Solve the exponential function by converting to a common base:

Question

Show step by step work:

27 ⁄ 9^6x+3×81^13x= 81^x+1⁄ 27×9^-7x-52

Answer 1

(27 / (9^6x+3)) * (81^13x) = (81^x+1) / (27 * 9^7x-52)

I recommend writing out these steps to follow along more accurately and to learn how to do problems like in the future.

We first want to isolate the common bases to their respective sides. Clearly 81^x+1 should be on the same side of the equation as 81^13x so that we may simplify.

Divide both sides by 81^13x and you end up with:

(27 / (9^6x+3)) = (81^x+1) / [(27 * 9^7x-52) * (81^13x)]

Now we multiply both sides by (27 * 9^7x-52) and you end up with:

(27² * 9^7x-52) / (9^6x+3)) = (81^x+1) / (81^13x)

Now let's focus on the left side of the equation.

We know that in order to simplify further we require isolating some amount of 9^7x-52 so that we can cancel out with 9^6x+3

When we talk about exponents, it is key to remember the basic principles of how exponents work when figuring out how to break down functions that use entire expressions with variables as an exponent.

We know that when a¹ * a¹ is written as an exponent, we are adding together the values of the exponents.

a¹⁺¹ = a²

This holds true across all exponents.

Therefore a^f(x) * a^g(x) = a^{f(x) + g(x)}

^{Now lets use this rule to analyze the function we are working with.}

a^{f(x) + g(x)}

Where:

a = 9

f(x) = 7x

g(x) = -52

9^7x-52 = 9^7x * 9^-52

But this breakdown is not helpful in simplifying the equation above. Instead we want to break 9^7x-52 into parts that include 9^6x+3 so we can cancel out.

9^7x-52 = 9^6x * 9^x * 9³ * 9^-55

This function is true because we know that:

6x + x + 3 - 55 = 7x - 52

Now we combine the broken down parts into useful parts:

9^7x-52 = 9^6x * 9^x * 9³ * 9^-55 = 9^6x+3 * 9^x * 9^-55

Now lets go back to the original equation and plug this in.

(27² * 9^7x-52) / (9^6x+3)) = (81^x+1) / (81^13x)

Substitute the new broken down equation for 9^7x-52 with 9^6x+3 * 9^x * 9^-55:

(27² * 9^6x+3 * 9^x * 9^-55) / (9^6x+3)) = (81^x+1) / (81^13x)

Now we can cancel out 9^6x+3 from the equation.

(27² * 9^x * 9^-55) = (81^x+1) / (81^13x)

We will now do the same process with 81^x+1:

81^x+1 = 81^13x * 81 * 81^-12x

We can double check that our breakdown was correct by adding together all the exponents, and it should come out = x + 1

13x + 1 + (-12x) = x + 1

Now we plug in the substitution for 81^x+1 with 81^13x * 81 * 81^-12x :

(27² * 9^x * 9^-55) = (81^13x * 81 * 81^-12x) / (81^13x)

Now we can cancel out 81^13x

(27² * 9^x * 9^-55) = ( 81 * 81^-12x)

Now that the hard part is out of the way, we can whip out our calculators to solve for 27²

729 * 9^x * 9^-55 = 81 * 81^-12x

Now divide both sides by 81.

9 * 9^x * 9^-55 = 81^-12x

Now we can combine 9 * 9^-55 to give us:

9^x * 9^-54 = 81^-12x

Now comes another useful rule of exponents.

(a^x)^x = a^{x * x}

This means that 81^-12x = (9²)^-12x = 9^-24x

Now we may substitute this value for 81^-12x as it allows us to simplify further with the rest of the equation being in base 9^x:

9^x * 9^-54 = 9^-24x

The final rule we will be using to help us simplify even further is very similar to the first rule we talked about earlier.

a^f(x) / a^g(x) = a^{f(x) - g(x)}

This means we can divide both sides of the equation by 9^-24x .

9^x * 9^-54 / 9^-24x = 9^-24x / 9^-24x

The right side of the equation cancels out to 1 for:

9^x * 9^-54 / 9^-24x = 1

Now we combine exponents with the rule mentioned above:

9^x / 9^-24x = 9^{x - (-24x)} = 9^25x

So the equation can now substitute this value for 9^x / 9^-24x so the equation now reads:

9^25x * 9^-54 = 1

Divide both sides by 9^-54

9^25x = 1 / 9^-54 = 9⁵⁴

Now we dive into a rule involving square roots.

sqrt ^g(x)(a^f(x)) = a^{f(x) / g(x)}

So we will take a g(x) root where g(x) = 25,

9^{25x / 25} = 9^{54 / 25}

This simplifies out to:

9^x = 9^54/25

To solve for x, we must now write this equation in logarithmic form.

We know that for logarithmic functions:

a^x = b

Log_ab = x

So for our equation:

x = Log₉9^54/25

This can be a final solution, but another way to write this is:

x = Ln (9^54/25) / Ln (9)

And if you put that into your calculator your final value for x should come out to approximately:

x = 4.745 / 2.197

x = 2.160

(approximation rounded to the nearest 3rd decimal place)

Answer 2

Quickly Reviewing the Properties of exponents..........

(I) B^m * B^n = B^(m+n)

(II) B^m / N^n = B^(m-n)

(III) (B^m)^n = B^(m*n) = B^(mn)

Then, there is the issues of knowing and recognizing the powers

of integers, so as to identify the common base

In this problem, the common base is 3:

3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6= 729

Writes the problem like this, using the dashes as the fraction bar:

13x x+1 -7x-52

27 81 81 9

-------- * ---------- = --------------- * ----------------

6x+3

9 1 27 1

Multiplying the fractions:

13x x+1 (-7x-52)

(27) (81) (81) 9

----------------- = ---------------------

6x+3

9 27

The common base is 3: 3^2 = 9, 3^3 = 27, 3^4 = 81

3 4 (13x) 4 (x+1) 2 (-7x-52)

(3 )( 3 ) (3 ) (3 )

---------------------- = ------------------------------

2 (6x+3) 3

(3 ) (3 )

(52x + 3) 4(x+1) + 2(-7x-52)

3 3

------------------------- = -----------------------

(12x + 6) 3

3 3

(52x + 3 ) - (12x+6) = (4x+4 + -14x - 104-3)

3 3

(52x + 3 - 12x - 6) (4x + 4 -14x - 104-3)

3 = 3

(40x - 3) (-10x - 103)

3 = 3

40x - 3 = -10x - 103

50x - 3 = -103

50x = -100

x = -2