10^n+2= 1/10000 ^1-n

Question

 2^x+^2 +3 • 2^x=56,

2^x+1• 2^3-2x =8^2-x

(4/9) ^m+2 = (8/27)^1+m

 

10^n+2= 1/10000 ^1-n

 

please help me to solve this question ,thank you

Answer 1

Exponential terms with the same base and exponent can be treated as if they were variables for the purpose of adding or subtracting like terms.

 

2^x + 2 + 3 * 2^x = 56

2^x + 3 * 2^x = 54

4 * 2^x = 54

2^x = 27/2

 

To get the variable out of the exponent, convert to log form

x = log₂ (27/2)

x = log₂ 27 - log₂ 2

x = log₂ 3³ - 1

x = 3 log₂ 3 - 1

 

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If you can get two exponential expressions to have the same base and be equal, then their exponents must also be equal.

(4/9)^m+2 = (8/27)^1+m

((2/3)²)^m+2 = ((2/3)³)^m+1

(2/3)^2m+4 = (2/3)^3m+3

2m+4 = 3m+3

1 = m

 

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Again, you can manipulate the expressions to have the same base, and set the exponents equal. Remember that 1/x^a = x^-a

10ⁿ⁺² = 1/10,000^1-n

10ⁿ⁺² = 10,000^-1+n

10ⁿ⁺² = (10⁴)^n-1

10ⁿ⁺² = 10^4n-4

n+2 = 4n - 4

6 = 3n

2 = n