7^x+2 +2*7^x-1=375 2^2x-9<1 log в осн3 27-log в осн9 81

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вопрос

To solve the first equation, we need to simplify it:

7^x+2 + 27^x-1 = 375
7^x 7^2 + 2 (7^x / 7) = 375
7^x 49 + 2 (7^x / 7) = 375
49 7^x + 2 (7^x / 7) = 375
49 7^x + 14/7 7^x = 375
49 7^x + 14 7^x = 375
63 7^x = 375
7^x = 375 / 63
7^x = 5

Now, we need to solve for x using logarithms:

log7(7^x) = log7(5)
x = log7(5)

For the second equation, simplify it first:

2^(2x-9) < 1
2^(2x) 2^(-9) < 1
2^(2x) (1/2^9) < 1
2^(2x) * (1/512) < 1
2^(2x) < 512

Now, 512 = 2^9, so the inequality becomes:

2^(2x) < 2^9
2x < 9
x < 4.5

Therefore, the solution to the second equation is x < 4.5.

For the third equation, we have:

log3(27) - log9(81)
log3(3^3) - log9(9^2)
3log3(3) - 2log9(3)
31 - 2(1/2)
3 - 1
2

Therefore, log3(27) - log9(81) = 2.