Solving exponential equations algebraically?

In this case the bases are powers of 3. Notice that:

(1/9)^(x-3) = 3(27)^(x-3)

(3^-2)^(x-3) = 3(3^3)^(x-3)

(3^-2)^(x-3) = 3(3)^(3*(x-3))

3^(-2*(x-3)) = 3^(3*(x-3) + 1)--------------> from this step to the next is like taking the log_3 in both sides

-2(x-3) = 3*(x-3) +1

-2x+6 = 3x-9 +1

6+9-1 = 3x+2x

14=5x

x = 14/5

Solve The Exponential Equation Algebraically

When you see common bases or common exponents, those are cues for you to use shortcuts.

Let x-3 = y

1(1/9)^y = 3(27)^y

1 = 3[27^y/(1/9)^y]

1 = 3^(5y+1)

y = -1/5

x = y +3 or 2 4/5 or 2.8

Your goal is to get a common base to the exponents on both sides: Currently you have 1/9 and 27 which are both powers of 3.

Specifically:

1/9 = 3^(-2)

27 = 3^3

So you have:

(3^(-2))^(x-3) = 3 * (3^3)^(x-3)

Next, remember this rule of exponents:

(a^b)^c = a^(bc)

So we can rewrite this as:

3^(-2(x-3)) = 3^1 * 3^(3(x-3))

Finally, remember this rule:

a^b * a^c = a^(b + c)

Use this to combine the right side:

3^(-2(x-3)) = 3^(1+3(x-3))

We now have both sides with the same format of

3^a = 3^b

So cancel the 3 from both sides (equivalently, take the log base 3 of both sides):

-2(x-3) = 1 + 3(x-3)

Expand:

-2x + 6 = 1 + 3x - 9

Combine like terms:

6 - 1 + 9 = 3x + 2x

14 = 5x

x = 14/5

x = 2.8

First, note that 1/9 = 3^(-2), and 27 = 3^3.

So now we have

(3^-2)^(x-3) = 3*(3^3)^(x-3)

3^(6 - 2x) = 3*3^(3x - 9) = (3^1)(3^(3x - 9)) = 3^(3x - 8) Therefore

6 - 2x = 3x - 8

I know you can finish it now!

You can use either log or ln

ln[(1/9)^(x-3)] = ln[3(27)^(x-3)]

--> (x-3)ln(1/9) = ln(3) + (x-3)ln(27)

--> (x-3)[ln(1) - ln(9)] = ln(3) + (x-3)ln(3*9)

--> (x-3)[ln(1) - ln(9)] = ln(3) + (x-3)[ln(3) + ln(9)]

--> (x-3)ln(1) - (x-3)ln(9) = ln(3) + (x-3)ln(3) + (x-3)ln(9)

--> (x-3)(0) - 2(x-3)ln(3) = ln(3) + (x-3)ln(3) + 2(x-3)ln(3) Note: 9 = 3^2, ln(3^2) = 2ln(3)

--> -5(x-3)ln(3) = ln(3)

--> -5(x-3) = 1

--> -5x + 15 = 1

--> -5x = -14

--> x = 14/5 = 2.8

Can you tell what's ^ this stand for then I can easily solve