# Determine the exact value of x for `(1/8)^(x - 3) = 2*16^(2x + 1)` .

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The equation to be solved is: `(1/8)^(x - 3) = 2*16^(2x + 1)`

`(1/8)^(x - 3) = 2*16^(2x + 1)`

=> `((1/8)^x)/(1/8)^3 = 2*(2^4)^(2x + 1)`

=> `(8^3)/(8^x) = 2*2^(8x)*2^4`

=> `2^9/(8^x) = 2^5*2^(8x)`

=> `2^4 = 2^(8x)*2^(3x)`

=> `2^4 = 2^(11x)`

As the base is the same equate the exponent.

=> `11x = 4`

=> `x = 4/11`

**The solution of the equation `(1/8)^(x - 3) = 2*16^(2x + 1) ` is `x = 4/11` **

For problem #1, `3^(2x) = 5(3^x) +36`

Solution:

>> Make the right side equal to zero.

`3^(2x) - 5(3^x) - 36 = 0`

>> Express the above expression in a quadratic form.

To do this replace 3^x with another variable.

Let, `3^x = z`

`(3^x)^2 - 5(3^x) - 36 = 0`

`z^2 - 5z - 36 = 0`

>> Factor left side.

`(z -9)(z + 4) = 0`

>> Set each factoe equal to zero.

`(z- 9) = 0` and `(z + 4)= 0`

` z = 9` z = -4

>> Note that 3^x can not be equal to -4. Note that for a positive base, the resulting value of an exponential expression is always positive. Hence, we will only consider the positive value of z, to solve for x.

`3^x = 9`

>> Express 9 with its prime factors. *(Note that 9 = 3*3 = 3^2.*)

`3^x = 3^2`

>> Since both sides have the same base, set the exponents equal to each other.

`x = 2`

**Hence. the value of x is 2.**

For problem #2, `(1/8)^(x-3) = 2(16^(2x+1))`

Solution:

>> Express the numbers 8 and 16 with its prime factors. *(Note 8=2*2*2= 2^3 and 16 = 2*2*2*2=2^4 )*

`(1/2^3)^(x-3) = 2(2^((4)(2x+1)))`

` `>> Then, re-write left side using the rule for negative exponents. Rule is 1/a^m = a^-m.

`2^((-3)(x-3)) = 2(2^((4)(2x+1)))`

>> Use distributive property to simplify the exponents.

`2^(-3x+9) = 2 (2^(8x+4))`

`2^(-3x+9) = 2^1(2^(8x+4))` *Note that* `2 = 2^1`

>> Simplify right side by using the rule of multiplication of same base, which is a^m * a^n = a^(m+n) .

`2^(-3x+9) =2 ^(1+8x+4)`

`2^(-3x+9) = 2^(8x+5)`

>> Since the both sides have the same base, set the exponents equal to each other to solve for x.

`-3x+9 = 8x + 5`

`-11x = -4`

`x = 4/11`

` `

**Hence, `x = 4/11` .**

For problem #3, `3^(x^2+20) = (1/27)^(3x)`

Solution:

>> Express 27 with its prime factors. ( 27 = 3*3*3 = 3^3).

`3^(x^2 + 20) = (1/3^3)^(3x)`

>> Re-write (1/3^3) using the rule for negative exponent. Rule is 1/a^m = a^-m .

`3^(x^2 + 20) = 3^((-3)(3x))`

`3^(x^2 + 20) = 3^(-9x)`

>> Since both sides have same base, set the exponents equal to each other in order to solve for x.

`x^2 + 20 = -9x`

` ` `x^2 + 9x + 20 = 0`

>> Factor left side.

`(x+4)(x+5) = 0`

>> Set each factor equal to zero.

`(x+4) =0` and `(x+5) = 0`

`x = -4` and `x = -5`

**Hence, the value of x are -4 and -5. **