Exponential Equations

Exponential Equations: Solutions Without Logs

In this section, we will explore various exponential equations and methods to solve for the variable without necessarily resorting to logarithms when possible.

Example 1: Solving an Exponential Equation with a Change of Base

Consider the equation:
$3^{x + 2} = 9^{2x - 3}$
To solve for the value of $x$ without using logarithms, we first convert the base of 9 into a power of 3. Since $9 = 3^2$, we can rewrite it:
$3^{x + 2} = (3^2)^{2x - 3}$
Using the rules of exponents, specifically that $(a^m)^n = a^{mn}$, we get:
$3^{x + 2} = 3^{(2)(2x - 3)} = 3^{4x - 6}$
Since the bases are equal, we can equate the exponents:
$x + 2 = 4x - 6$
Now, solve for $x$:

• Subtract $x$ from both sides:
  $2 + 6 = 4x - x$

• This simplifies to:
  $8 = 3x$

• Therefore:
  $x = rac{8}{3}$
  Thus, the value of $x$ for this equation is $rac{8}{3}$.

Example 2: Finding x from Different Exponential Bases

Now, consider the equation:
$8^{4x - 12} = 16^{5x - 3}$
We will convert both sides to use a common base, which in this case is 2.
Given that:

• $8 = 2^3$

• $16 = 2^4$
  We can rewrite the equation as follows:
  $(2^3)^{4x - 12} = (2^4)^{5x - 3}$
  Following the exponent rules gives:
  $2^{12x - 36} = 2^{20x - 12}$
  Since the bases are equivalent, we equate the exponents:
  $12x - 36 = 20x - 12$
  Now, manipulate the equation:

• Subtract $12x$ from both sides:
  $-36 = 8x - 12$

• Add $12$:
  $-36 + 12 = 8x$
  This yields:
  $-24 = 8x$
  Hence, $x = -3$
  This indicates the solution for $x$ is $-3$.

Example 3: Solving with Different Powers of 3

Examining the equation:
$27^{3x - 2} = 81^{2x + 7}$
Here, we can once again convert to base 3:

• $27 = 3^3$

• $81 = 3^4$
  Reformulating gives:
  $(3^3)^{3x - 2} = (3^4)^{2x + 7}$
  Thus, the equation becomes:
  $3^{9x - 6} = 3^{8x + 28}$
  Now we can set the exponents equal:
  $9x - 6 = 8x + 28$
  Rearranging the equation involves:

• Subtracting $8x$ from both sides:
  $x - 6 = 28$

• Adding $6$ results in:
  $x = 34$
  Hence, the solution is $34$.

Example 4: Using Natural Logarithms

For the equation $3^x = 8$, we can find $x$ by applying logarithms:
Taking the logarithm of both sides, we have:
$ext{log}(3^x) = ext{log}(8)$
Applying the power rule transforms to:
$x ext{log}(3) = ext{log}(8)$
Thus, solving for $x$ yields:
$x = rac{ ext{log}(8)}{ ext{log}(3)}$
Calculating this gives an approximate value of $x ext{ is about } 1.8928$.
You can verify this by checking if $3^{1.8928} ext{ approximates } 8$.

Example 5: Another Application of Natural Logarithm

Consider the equation $e^x = 7$.
Taking the natural logarithm gives us:
$ext{ln}(e^x) = ext{ln}(7)$
Since $ext{ln}(e)= 1$, we find:
$x = ext{ln}(7)$, with an approximate value of $x ext{ being } 1.9459$.

Example 6: Solving Mixed Exponential Equations

Explore the equation $5 + 4^{x - 2} = 23$:
We will first isolate the exponential part:

• Subtracting 5 gives:
  $4^{x - 2} = 18$
  We can't convert 18 into a base of 4, thus we take logarithms:
  $ext{log}(4^{x - 2}) = ext{log}(18)$
  Utilizing the power property yields:
  $(x - 2) ext{log}(4) = ext{log}(18)$
  Dividing both sides gives:
  $x - 2 = rac{ ext{log}(18)}{ ext{log}(4)}$
  From this we can solve for $x$ as:
  $x = 2 + rac{ ext{log}(18)}{ ext{log}(4)}$
  Numerically, this results in approximately $4.08496$ as the solution for $x$.

Example 7: Solving an Equation with e

For the equation $3 + 2e^{3 - x} = 7$, we start by isolating the exponential:

• Subtracting 3:
  $2e^{3 - x} = 4$

• Dividing by 2 gives:
  $e^{3 - x} = 2$
  Now, apply the natural logarithm:
  $ext{ln}(e^{3 - x}) = ext{ln}(2)$
  This simplifies to:
  $3 - x = ext{ln}(2)$
  So reordering gives us:
  $x = 3 - ext{ln}(2)$, which is approximately $2.3069$.

Example 8: Factors and Exponentials

Examining the equation $3^{x^2 + 4} = rac{1}{27}$ leads us to recognize that:
Since $27 = 3^3$, we can rewrite:
$rac{1}{27} = 3^{-3}$
Thus we have:
$3^{x^2 + 4} = 3^{-3}$
This allows us to equate exponents:
$x^2 + 4 = -3$
Rearranging gives:
$x^2 + 7 = 0$
Factoring yields:
$(x + 3)(x + 1) = 0$
The solutions are $x = -3$ and $x = -1$.

Example 9: Adding Exponents in a Common Base

For the equation $2^{x^2} imes 2^{3x} = 16$, we rewrite 16 as:
$16 = 2^4$
Using properties of exponents gives:
$2^{x^2 + 3x} = 2^4$
We can equate exponents:
$x^2 + 3x = 4$
Rearranging into standard form provides:
$x^2 + 3x - 4 = 0$
Factoring yields:
$(x + 4)(x - 1) = 0$
Thus, the solutions are $x = -4$ and $x = 1$.

Example 10: Factoring Exponential Equations

In the case of $4^{2x - 20} imes 4^{x + 64} = 0$, we substitute:
Letting $a = 4^x$, we find that:

• So, $4^{2x} = a^2$ and $4^x = a$
  This allows us to set the equation as:
  $a^2 - 20a + 64 = 0$
  Factoring gives:
  $(a - 16)(a - 4) = 0$
  Considering that $a = 4^x$:

• Therefore $4^x = 16$ implies $x = 2$

• Meanwhile, $4^x = 4$ gives $x = 1$.
  Thus, the solutions are $x = 2$ and $x = 1$.

Example 11: Finding Roots from Expanding Exponential Expressions

Lastly, let's solve $3^{2x} - 3^{2x - 1} = 18$.
Factoring out the common term leads to:
$3^{2x}(1 - rac{1}{3}) = 18$
This highlights our equation becoming $rac{2}{3}3^{2x} = 18$.
Multiplying by $rac{3}{2}$ gives:
$3^{2x} = 27$
Recognizing that $27 = 3^3$, we can equate exponents:
$2x = 3$
Thus, $x = rac{3}{2}$.

Each section showcases methods of solving exponential equations by converting bases, utilizing logarithms when necessary, factoring, or using algebraic manipulations without extensive recomputation or loss of detail.