2*16^x-3*2^4x-1+7*4^2x-2=120

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вопрос

To solve the equation 216^x - 32^4x - 1 + 7*4^2x - 2 = 120, we first need to simplify it using the properties of exponents and basic arithmetic operations.

We know that:
16 = 2^4
4 = 2^2

Therefore, we can rewrite the equation as:
2(2^4)^x - 32^(4x) - 1 + 7(2^2)^(2x) - 2 = 120
22^(4x) - 32^(4x) - 1 + 72^(4x) - 2 = 120
6*2^(4x) - 3 = 120

Now, let's rewrite 62^(4x) as (23)2^(4x) = 622^(4x) = 12(2^(4x)). Now our equation looks like:

12*2^(4x) - 3 = 120

Now we can add 3 to both sides of the equation to isolate the term with the exponent:

12*2^(4x) = 123

Divide both sides by 12:

2^(4x) = 123/12
2^(4x) = 10.25

Now we can rewrite 10.25 as 41/4:

2^(4x) = 41/4

To solve the equation above, we take the natural logarithm of both sides:

ln(2^(4x)) = ln(41/4)
4x * ln(2) = ln(41/4)

Now divide by ln(2) to solve for x:

4x = ln(41/4) / ln(2)
x = (ln(41/4) / ln(2)) / 4

After calculating the result, we will obtain the value of x.