Math Problem Statement

8^(1/3 x+2)=2^(x+6) 36^(1/2 x+4)=6^(x+8) 2^x+2^(x+1)=3⋅2^x

Solution

Let's solve these equations one by one.

Equation 1:

8(13x+2)=2x+68^{\left(\frac{1}{3} x + 2 \right)} = 2^{x + 6}

Step 1: Express both sides in terms of base 2.

  • Recall that 8=238 = 2^3, so we can write: 8(13x+2)=(23)(13x+2)=23(13x+2)=2x+68^{\left(\frac{1}{3} x + 2 \right)} = (2^3)^{\left(\frac{1}{3} x + 2 \right)} = 2^{3\left(\frac{1}{3} x + 2 \right)} = 2^{x + 6}

Step 2: Equating the exponents (since the bases are the same): 3(13x+2)=x+63\left(\frac{1}{3} x + 2 \right) = x + 6

Simplifying the left-hand side: x+6=x+6x + 6 = x + 6

This results in a true statement, meaning the equation holds for all values of xx. So, the solution is all real numbers.

Equation 2:

36(12x+4)=6x+836^{\left(\frac{1}{2} x + 4 \right)} = 6^{x + 8}

Step 1: Express both sides in terms of base 6.

  • Recall that 36=6236 = 6^2, so we can write: 36(12x+4)=(62)(12x+4)=62(12x+4)=6x+836^{\left(\frac{1}{2} x + 4 \right)} = (6^2)^{\left(\frac{1}{2} x + 4 \right)} = 6^{2\left(\frac{1}{2} x + 4 \right)} = 6^{x + 8}

Step 2: Equating the exponents (since the bases are the same): 2(12x+4)=x+82\left(\frac{1}{2} x + 4 \right) = x + 8

Simplifying the left-hand side: x+8=x+8x + 8 = x + 8

This is another true statement, meaning this equation also holds for all values of xx. So, the solution is all real numbers.

Equation 3:

2x+2x+1=32x2^x + 2^{x+1} = 3 \cdot 2^x

Step 1: Factor out 2x2^x from the left-hand side: 2x(1+2)=32x2^x(1 + 2) = 3 \cdot 2^x

This simplifies to: 2x3=32x2^x \cdot 3 = 3 \cdot 2^x

Step 2: Divide both sides by 2x2^x (assuming 2x02^x \neq 0): 3=33 = 3

This is trivially true, meaning the equation holds for all values of xx. So, the solution is all real numbers.

Final Answer:

For all three equations:

  • The solutions are all real numbers.

Would you like more detailed steps on any of these equations, or is there a specific concept you want to explore further?

Follow-up Questions:

  1. How do you approach solving exponential equations with different bases?
  2. How can we solve exponential equations involving fractional exponents?
  3. What is the general method for solving equations where exponents are linear expressions?
  4. What strategies can be used to solve equations with different logarithmic bases?
  5. How do we solve exponential equations when the bases cannot be easily rewritten in terms of each other?

Tip:

When dealing with exponential equations, always look for ways to rewrite both sides using the same base, as this simplifies the process of equating the exponents.

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Math Problem Analysis

Mathematical Concepts

Exponential Equations
Algebra
Exponents
Logarithms

Formulas

a^(m+n) = a^m * a^n
a^(m) = b^(n) implies m = n when a = b
Factoring expressions
Equating exponents for like bases

Theorems

Properties of exponents
Logarithmic and exponential equivalence

Suitable Grade Level

Grades 8-10