A 3x-1 2x-9 1) ; 2) 2×2-1 - 3× 9x2-1 9x2+2. 5) 4 3) 2x2+x-6 - 2x2+x-9 = 56; 4) 25* - 6.5* + 5 = 0; 5) 92-1 - 36 3*2-3 + 3 = 0; 6) 9*+1 - 3 3*+3 - 27 3*-3 + 27 = 0; 7) 15 3* - 2 3*; 8) 2* +10 9 4 2t-2; 9) 101+x2 - 101-x2 = 99; 10) ln(3x - 5) = 0; 11) 2 log0,5x = log0,5(2x2 - x); 12) log2(4 - x) + log2(1 - 2x) = 2 log23; 13) ln(x2 - 6x + 9) = ln3 + ln(x + 3); 14) 2log3 x - 7log3 x + 3 = 0; 31gx+19 15) = 2 lg x + 1. 3lgx - 1

Thanks for the list. The problems are a bit garbled (especially items 1–9 and 16+). I can solve clearly stated items, and I can help you if you can re-send the rest in a clean format. Here are the solvable ones I can extract (10–15), with brief steps and answers: 10) ln(3x − 5) = 0 - Domain: 3x − 5 > 0 → x > 5/3 - ln(s) = 0 ⇒ s = 1 - 3x − 5 = 1 ⇒ x = 2 - Check: 3·2 − 5 = 1 > 0. Solution: x = 2 11) 2 log base 0.5 of x = log base 0.5 of (2x^2 − x) - Let base b = 0.5 (valid since b > 0 and b ≠ 1) - Use property: 2 log_b x = log_b(x^2) So log_b(x^2) = log_b(2x^2 − x) - Since log_b is one-to-one, x^2 = 2x^2 − x ⇒ 0 = x^2 − x ⇒ x(x − 1) = 0 - Domain: x > 0 and 2x^2 − x > 0 - Solutions from equation: x = 0 or x = 1; x = 0 is not allowed (x > 0), so x = 1 - Check: 2x^2 − x = 2 − 1 = 1 > 0 - Solution: x = 1 12) log_2(4 − x) + log_2(1 − 2x) = 2 log_2 3 - Use log addition: log_2[(4 − x)(1 − 2x)] = log_2(3^2) = log_2 9 - Domain: 4 − x > 0 → x < 4; 1 − 2x > 0 → x < 1/2 - Solve (4 − x)(1 − 2x) = 9 2x^2 − 9x + 4 = 9 → 2x^2 − 9x − 5 = 0 Discriminant: Δ = 81 + 40 = 121 → x = [9 ± 11]/4 x = 5 or x = −1/2 - Check domain: x must be < 1/2, so x = −1/2 is valid; x = 5 is not - Solution: x = −1/2 13) ln(x^2 − 6x + 9) = ln 3 + ln(x + 3) - Left: ln[(x − 3)^2]; right: ln[3(x + 3)] - Exponentiate: (x − 3)^2 = 3(x + 3) ⇒ x^2 − 6x + 9 = 3x + 9 x^2 − 9x = 0 ⇒ x(x − 9) = 0 - Domain: (x − 3)^2 > 0 (x ≠ 3) and x + 3 > 0 (x > −3) - Solutions: x = 0, x = 9 (both satisfy domain) - Solutions: x ∈ {0, 9} 14) 2 log_3 x − 7 log_3 x + 3 = 0 - This is (2 − 7) log_3 x + 3 = 0 ⇒ −5 log_3 x + 3 = 0 - log_3 x = 3/5 ⇒ x = 3^(3/5) - Solution: x = 3^(3/5) 15) If the problem is 2 lg x + 1 = 3 lg x − 1 - Bring terms: (2 − 3) lg x = −1 − 1 ⇒ −lg x = −2 - lg x = 2 ⇒ x = 10^2 = 100 - Solution: x = 100 If you intended different forms for items 1–9 and 16+, please re-send those problems clearly (preferably with parentheses and using consistent notation, or attach an image). I’ll solve all of them step by step. Tips for tackling the rest (once you re-send them): - For rational equations, check domain restrictions where denominators cannot be zero and where expressions are defined. - For logarithmic equations, ensure all arguments are positive; use log identities to combine or equate arguments. - When you have equations mixing different bases (e.g., log_a and log_b), convert to a common base or use exponentiation carefully to avoid extraneous roots. - Always verify potential solutions in the original equation, due to domain restrictions.