Lg(7x-9)^2+lg(3x-4)^2=2

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вопрос

To solve the given equation, we first need to expand the squares in the logarithms.

Using the property that log(a^b) = b*log(a), we get:

2log((7x-9) (7x-9)) + 2log((3x-4) (3x-4)) = 2

Further simplifying, we get:

2log((7x-9)^2) + 2log((3x-4)^2) = 2

Now, we can use the property that log(a) + log(b) = log(ab) to combine the logarithms:

log((7x-9)^2 * (3x-4)^2) = 2

Taking 10 to the power of both sides, we get:

(7x-9)^2 * (3x-4)^2 = 10^2

Expanding both sides, we get:

(49x^2 - 126x + 81) * (9x^2 - 24x + 16) = 100

Expanding further, we get:

441x^4 - 1323x^3 + 1069x^2 - 2016x + 1296 = 100

Rearranging terms and simplifying, we get:

441x^4 - 1323x^3 + 1069x^2 - 2016x + 1196 = 0

Unfortunately, this is a quartic equation and cannot be solved easily by hand. You may need to use numerical methods or a graphing calculator to find the solutions for x.