Question

3.3^(2x+1)-103^x+1=0 need value of x

Answer 1

The value of
`x` is approximately -1.531.

Explanation:

Let
`3.3^(2\cdot x + 1)-103^(x+1) = 0`, we proceed to solve this expression by algebraic means:

1)
`3.3^(2\cdot x + 1)-103^(x+1) = 0` Given

2)
`3.3^(2\cdot x)\cdot 3.3 -103^(x)\cdot 103 = 0`
`a^(b)\cdot a^(c) = a^(b+c)`

3)
`(3.3^(x))^(2)\cdot 3.3 -\left[\left( √(103) \right)^(2)\right]^(x)\cdot 103 = 0`
`(a^(b))^(c) = a^(b\cdot c)`

4)
`(3.3^(x))^(2)\cdot 3.3 - \left[\left(√(103)\right)^(x)\right]^(2)\cdot 103 = 0`
`(a^(b))^(c) = a^(b\cdot c)`/Commutative property

5)
`\left[\left((3.3)/(√(103))\right)^(x)\right] ^(2)-(103)/(3.3) = 0` Existence of multiplicative inverse/Definition of division/Modulative property/
`a^(b)\cdot a^(c) = a^(b+c)`

6)
`\left((3.3)/(√(103)) \right)^(2\cdot x)=(103)/(3.3)` Existence of additive inverse/Modulative property/
`(a^(b))^(c) = a^(b\cdot c)`

7)
`\log \left((3.3)/(√(103)) \right)^(2\cdot x)=\log (103)/(3.3)` Definition of logarithm.

8)
`2\cdot x\cdot \log \left((3.3)/(√(103)) \right)= \log (103)/(3.3)`
`\log_(b) a^(c) = c\cdot \log_(b) a`

9)
`2\cdot x \cdot [\log 3.3-\log √(103)] = \log 103 - \log 3.3`
`\log_(b) (a)/(d)`

10)
`x\cdot (2\cdot \log 3.3-\log 103) = \log 103 - \log 3.3`
`\log_(b) a^(c) = c\cdot \log_(b) a`/Associative property

11)
`x = (\log 103-\log 3.3)/(2\cdot \log 3.3-\log 103)` Existence of multiplicative inverse/Definition of division/Modulative property

12)
`x \approx -1.531` Result

The value of
`x` is approximately -1.531.