Solve the following logarithmic and exponential equations

Question

1) 3^{3x + 2} = 81^2x-2

2) x·2^x - x² = 135

3) log(x+4) = 2

4)log₃(x² + 6x) = 3

5) e^{5x - 15} = 16

Answer 1

1) 3^{3x + 2} = 81^{2x - 2}

3^{3x + 2} = (3⁴)^{2x - 2}

Note: (b^a)ⁿ = b^an

3^{3x + 2} = 3^{8x - 8}

3x + 2 = 8x - 8 (same base; set the exponents equal to each other)

-5x = -10

x = 2

Check: 3^{3(2) + 2} = 81^{2(2) - 2}

3⁸ = 81²

6561 = 6561

2) x•2^x - x² = 135

x(2^x - x) = 135

Note: Find two factors whose product is equal to 135.

1, 135

3, 45

5, 27

9, 15

Guessing and testing with a small sample size will give you x = 5. Substitute x = 5 into the original equation to verify.

5 • 2⁵ - 5² = 135

5 • 32 - 25 = 135

160 - 25 = 135

3) log(x + 4) = 2

Note: log_b(m) = n if and only if bⁿ = m

10² = x + 4

100 = x + 4

96 = x

x = 96

Check: log(96 + 4) = 2

10² = 96 + 4

100 = 100

4) log₃(x² + 6x) = 3

Note: log_b(m) = n if and only if bⁿ = m

3³ = x² + 6x

0 = x² + 6x - 27

(x + 9)(x - 3) = 0

x + 9 = 0 or x - 3 = 0

x = -9 or x = 3

Domain of a log function: (0, ∞)

Test both answers

x = 3

log₃(3²+6(3)) = 3

3³ = 9 + 18

27 = 27

x = -9

log₃((-9)² + 6(-9)) = 3

3³ = 81 - 54

27 = 27

Answers: x = 3 or x = -9

5) e^{5x - 15} = 16

Note: log_b(m) = n if and only if bⁿ = m

If b = e, the log_e(m) = n becomes ln (m) = n. (The natural logarithm)

Or, take the natural logarithm to both sides to get:

ln(e^{5x - 15)} = ln(16)

5x - 15 = ln(16)

5x = 15 + ln(16)

x = (15 + ln(16))/5

or

x = 3 + (1/5)ln(16)

Answer 2

1) 3^{3x + 2} = 81^2x-2

3^{3x + 2} = 3^{4(2x -2)}

3x + 2 = 4(2x - 2)

3x + 2 = 8x - 8

5x = 10

x = 2

Is 81 a power of 3? Yes, it is. 81 = 3⁴. ALWAYS search for this initially.

Now that we have found an equivalent base on both sides, we can set the exponents equal to one another.

This becomes a simple linear equation - we can begin by distributing the 4 on the right-hand side.

Now we subtract 3x from both sides, and add 8 to both sides.

Dividing both sides by 5 gives us our final answer.

2) x·2^x - x² = 135

x(2^x - x) = 135

Set x = 5

5(2⁵ - 5) = 135

2⁵ - 5 = 27

x = 5

This one looks rather challenging, but testing factors makes it significantly easier.

We begin by factoring out an x on both sides.

Checking the factors of 135 yields (1, 3, 5, 9, 15, 27, 45, 135). After testing, setting x equal to 5 solves the equation, as illustrated next...

Substituting in 5 for x, we can divide both sides by 5 to get...2⁵ = 32, and 32 minus 5 indeed yields 27, so x does indeed equal 5.

The above question in particular highlights the fact that it's not always practical to solve complicated equations purely algebraically - some indeed require approximations, or testing that may involve factoring in order to effectively solve. In fact, many, many complicated equations are solved through approximations computationally.

3) log(x+4) = 2

10^{log(x + 4)} = 10²

x + 4 = 100

x = 96

This is a much simpler equation compared to the last two - we begin by having both raised as a power of 10 to eliminated the logarithm.

The 10 and the log on the left side cancel out, and 10² = 100.

ow this is a simple linear equation. Solving for x we have x = 96.

4) log₃(x² + 6x) = 3

3^{log(3)(x^2 + 6x)} = 3³

x² + 6x = 27

x² + 6x - 27 = 0

(x + 9)(x - 3) = 0

x = 3, x = -9

Not all logarithmic equations are log-base-10 or -base-e. You will sometimes see a base of an altogether different number.

Note: the log(3)(x^2 + 6x) expression on the right should be read as log₃(x² + 6x)

Cancelling the 3 with log₃ and simplifying 3³ to 27, this becomes quadratic.

Placing all values on the left side, do you see any way to factor this?

What adds to get 6 and multiplies to get -27? Factor out -27 and add together all complementary factors to find the one that works - in this case, 9 and -3.

5) e^{5x - 15} = 16

ln(e^{5x + 15}) = ln(16)

5x + 15 = ln(16)

5x = 15 - ln(16)

x = (15 - ln(16))/5

This looks intimidating at first glance, but it only requires one operation before it becomes a linear equation...

By taking the ln() of both sides, we can eliminate the exponential on the left.

Now we just solve this as a linear equation. For simplicity, you can leave the ln(16) as is, though some teachers may require you to put it in decimal form.

Placing all non-algebraic values on the right side allows us to solve for x.