Question

32×8∧(х-1)+3(4∧х+2∧х)=1

Answer 1

$32* 8^{x-1} +3*( 4^{x}+ 2^{x})=1, \frac{32* 8^{x} }{8}+3* 2^{2x}+3* 2^{x}-1=0, 4* 2^{3x}+3* 2^{2x}+3* 2^{x}-1=0, 2^{x}=z, 4 z^{3}+3 z^{2}+3z-1=0, (4z-1)( z^{2}+x+1)=0, 4z-1=0, z= \frac{1}{4} , 2^{x}= \frac{1}{4}, 2^{x}= 2^{-2}, x=-2.$

Answer 2

I believe there may be some ambiguity in the expression you provided.

In particular, it is not clear what the variable "x" represents, since it is not defined in the expression. It is also unclear whether the "^" symbol represents exponentiation or some other operation.

Assuming that "^" represents exponentiation and that the expression is meant to be solved for "x", we can rewrite it as follows:

32 × 8^(x-1) + 3(4^x + 2^x) = 1

We can simplify this expression by first using the distributive property to expand the second term:

32 × 8^(x-1) + 12 × 2^x + 6 × 4^x = 1

Next, we can combine the terms that have the same base:

32 × 2^3(x-1) + 12 × 2^x + 6 × 2^(2x) = 1

Now, we can use the fact that 2^3 = 8 to simplify the first term:

32 × 8^(x-1) = 256 × 2^(x-1)

Substituting this into the expression, we get:

256 × 2^(x-1) + 12 × 2^x + 6 × 2^(2x) = 1

We can simplify this further by factoring out a common factor of 2^x:

256 × 2^(x-1) + 12 × 2^x + 6 × 2^(2x) = 2^x

Dividing both sides by 2^x, we get:

256 × 2^(-1) + 12 + 6 × 2^(x) = 2^(-x)

Simplifying the left-hand side, we get:

128 + 6 × 2^(x) = 2^(-x)

Multiplying both sides by 2^x, we get:

128 × 2^x + 6 × 2^(2x) = 1

At this point, it is not clear whether this equation has a solution for x that can be expressed algebraically. It may be necessary to use numerical methods to find an approximate solution.