1). log5(x + 1) - log5(x - 1) = 22)?
1). log5(x + 1) - log5(x - 1) = 2 2). Logx + log(X - 3) = 1 3). Ln(x - 1) + ln(x + 2) = 1 4). Log(x ^ 2 - x - 2) = 2.
1). log5(x + 1) - log5(x - 1) = 2 2). Logx + log(X - 3) = 1 3). Ln(x - 1) + ln(x + 2) = 1 4). Log(x ^ 2 - x - 2) = 2.
Pregunta de Matemáticas · 1 respuesta · 8 votos · mejor respuesta de Valentina06
En resumen
1. <img src="https://tex.z-dn.net/?
1. <img src="https://tex.z-dn.net/?f=%5C%5C%5Clog_5%28x%2B1%29-%5Clog_5%28x-1%29%3D2%5C%5C%20x%2B1%3E0%5Cwedge%20x-1%3E0%5C%5C%20x%3E-1%20%5Cwedge%20x%3E1%5C%5C%20x%3E1%5C%5C%20%5Clog_5%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%3D2%5C%5C%205%5E2%3D%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%5C%5C%2025%3D%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%5C%5C%2025%28x-1%29%3Dx%2B1%5C%5C%2025x-25%3Dx%2B1%5C%5C%2024x%3D26%5C%5C%20x%3D%5Cfrac%7B26%7D%7B24%7D%5C%5C%20x%3D%5Cfrac%7B13%7D%7B12%7D%5C%5C" />0 \ wedge x - 1>0 \ \ x> - 1 \ wedge x>1 \ \ x>1 \ \ \ log_5 \ frac{x + 1}{x - 1} = 2 \ \ 5 ^ 2 = \ frac{x + 1}{x - 1} \ \ 25 = \ frac{x + 1}{x - 1} \ \ 25(x - 1) = x + 1 \ \ 25x - 25 = x + 1 \ \ 24x = 26 \ \ x = \ frac{26}{24} \ \ x = \ frac{13}{12} \ \ " alt = " \ \ \ log_5(x + 1) - \ log_5(x - 1) = 2 \ \ x + 1>0 \ wedge x - 1>0 \ \ x> - 1 \ wedge x>1 \ \ x>1 \ \ \ log_5 \ frac{x + 1}{x - 1} = 2 \ \ 5 ^ 2 = \ frac{x + 1}{x - 1} \ \ 25 = \ frac{x + 1}{x - 1} \ \ 25(x - 1) = x + 1 \ \ 25x - 25 = x + 1 \ \ 24x = 26 \ \ x = \ frac{26}{24} \ \ x = \ frac{13}{12} \ \ " align = "absmiddle" class = "latex - formula">
2.
<img src="https://tex.z-dn.net/?f=%5C%5C%5Clog%20x%2B%5Clog%28x-3%29%3D1%5C%5C%20x%3E0%20%5Cwedge%20x-3%3E0%5C%5C%20x%3E0%20%5Cwedge%20x%3E3%5C%5C%20x%3E3%5C%5C%20%5Clog%20x%28x-3%29%3D1%5C%5C%2010%5E1%3Dx%28x-3%29%5C%5C%2010%3Dx%5E2-3x%5C%5C%20x%5E2-3x-10%3D0%5C%5C%20x%5E2-5x%2B2x-10%3D0%5C%5C%20x%28x-5%29%2B2%28x-5%29%3D0%5C%5C%20%28x%2B2%29%28x-5%29%3D0%5C%5C%20x%3D-2%20%5Cvee%20x%3D5%5C%5C%20-2%5Cnot%3E3%5C%5C%20%5Cunderline%7Bx%3D5%7D" />0 \ wedge x - 3>0 \ \ x>0 \ wedge x>3 \ \ x>3 \ \ \ log x(x - 3) = 1 \ \ 10 ^ 1 = x(x - 3) \ \ 10 = x ^ 2 - 3x \ \ x ^ 2 - 3x - 10 = 0 \ \ x ^ 2 - 5x + 2x - 10 = 0 \ \ x(x - 5) + 2(x - 5) = 0 \ \ (x + 2)(x - 5) = 0 \ \ x = - 2 \ vee x = 5 \ \ - 2 \ not>3 \ \ \ underline{x = 5}" alt = " \ \ \ log x + \ log(x - 3) = 1 \ \ x>0 \ wedge x - 3>0 \ \ x>0 \ wedge x>3 \ \ x>3 \ \ \ log x(x - 3) = 1 \ \ 10 ^ 1 = x(x - 3) \ \ 10 = x ^ 2 - 3x \ \ x ^ 2 - 3x - 10 = 0 \ \ x ^ 2 - 5x + 2x - 10 = 0 \ \ x(x - 5) + 2(x - 5) = 0 \ \ (x + 2)(x - 5) = 0 \ \ x = - 2 \ vee x = 5 \ \ - 2 \ not>3 \ \ \ underline{x = 5}" align = "absmiddle" class = "latex - formula">
3.
<img src="https://tex.z-dn.net/?f=%5C%5C%5Cln%28x-1%29%2B%5Cln%28x%2B2%29%3D1%5C%5C%20x-1%3E0%20%5Cwedge%20x%2B2%3E0%5C%5C%20x%3E1%20%5Cwedge%20x%3E-2%5C%5C%20x%3E1%5C%5C%20%5Cln%28x-1%29%28x%2B2%29%3D1%5C%5C%20e%5E1%3D%28x-1%29%28x%2B2%29%5C%5C%20e%3Dx%5E2%2B2x-x-2%5C%5C%20x%5E2%2Bx-2-e%3D0%5C%5C%20%5CDelta%3D1%5E2-4%5Ccdot1%5Ccdot%28-2-e%29%5C%5C%20%5CDelta%3D1%2B8%2B4e%5C%5C%20%5CDelta%3D9%2B4e%5C%5C%20%5Csqrt%7B%5CDelta%7D%3D%5Csqrt%7B9%2B4e%7D%5C%5C%5C%5C%20x_1%3D%5Cfrac%7B-1-%5Csqrt%7B9%2B4e%7D%7D%7B2%7D%5C%5C%20x_1%3D-%5Cfrac%7B1%2B%5Csqrt%7B9%2B4e%7D%7D%7B2%7D%5C%5C%5C%5C%20x_2%3D%5Cfrac%7B-1%2B%5Csqrt%7B9%2B4e%7D%7D%7B2%7D%5C%5C%20x_2%3D-%5Cfrac%7B1-%5Csqrt%7B9%2B4e%7D%7D%7B2%7D%5C%5C%5C%5C%20-%5Cfrac%7B1%2B%5Csqrt%7B9%2B4e%7D%7D%7B2%7D%5Cnot%3E1%5C%5C%5C%5C%20%5Cunderline%7Bx%3D-%5Cfrac%7B1-%5Csqrt%7B9%2B4e%7D%7D%7B2%7D%7D%20" />0 \ wedge x + 2>0 \ \ x>1 \ wedge x> - 2 \ \ x>1 \ \ \ ln(x - 1)(x + 2) = 1 \ \ e ^ 1 = (x - 1)(x + 2) \ \ e = x ^ 2 + 2x - x - 2 \ \ x ^ 2 + x - 2 - e = 0 \ \ \ Delta = 1 ^ 2 - 4 \ cdot1 \ cdot( - 2 - e) \ \ \ Delta = 1 + 8 + 4e \ \ \ Delta = 9 + 4e \ \ \ sqrt{ \ Delta} = \ sqrt{9 + 4e} \ \ \ \ x_1 = \ frac{ - 1 - \ sqrt{9 + 4e}}{2} \ \ x_1 = - \ frac{1 + \ sqrt{9 + 4e}}{2} \ \ \ \ x_2 = \ frac{ - 1 + \ sqrt{9 + 4e}}{2} \ \ x_2 = - \ frac{1 - \ sqrt{9 + 4e}}{2} \ \ \ \ - \ frac{1 + \ sqrt{9 + 4e}}{2} \ not>1 \ \ \ \ \ underline{x = - \ frac{1 - \ sqrt{9 + 4e}}{2}} " alt = " \ \ \ ln(x - 1) + \ ln(x + 2) = 1 \ \ x - 1>0 \ wedge x + 2>0 \ \ x>1 \ wedge x> - 2 \ \ x>1 \ \ \ ln(x - 1)(x + 2) = 1 \ \ e ^ 1 = (x - 1)(x + 2) \ \ e = x ^ 2 + 2x - x - 2 \ \ x ^ 2 + x - 2 - e = 0 \ \ \ Delta = 1 ^ 2 - 4 \ cdot1 \ cdot( - 2 - e) \ \ \ Delta = 1 + 8 + 4e \ \ \ Delta = 9 + 4e \ \ \ sqrt{ \ Delta} = \ sqrt{9 + 4e} \ \ \ \ x_1 = \ frac{ - 1 - \ sqrt{9 + 4e}}{2} \ \ x_1 = - \ frac{1 + \ sqrt{9 + 4e}}{2} \ \ \ \ x_2 = \ frac{ - 1 + \ sqrt{9 + 4e}}{2} \ \ x_2 = - \ frac{1 - \ sqrt{9 + 4e}}{2} \ \ \ \ - \ frac{1 + \ sqrt{9 + 4e}}{2} \ not>1 \ \ \ \ \ underline{x = - \ frac{1 - \ sqrt{9 + 4e}}{2}} " align = "absmiddle" class = "latex - formula">
4.
<img src="https://tex.z-dn.net/?f=%5C%5C%5Clog%28x%5E2-x-2%29%3D2%5C%5C%20x%5E2-x-2%3E0%5C%5C%20x%5E2%2Bx-2x-2%3E0%5C%5C%20x%28x%2B1%29-2%28x%2B1%29%3E0%5C%5C%20%28x-2%29%28x%2B1%29%3E0%5C%5C%20x%5Cin%28-%5Cinfty%2C-1%29%5Ccup%282%2C%5Cinfty%29%5C%5C%2010%5E2%3Dx%5E2-x-2%5C%5C%20100%3Dx%5E2-x-2%5C%5C%20x%5E2-x-102%3D0%5C%5C%20%5CDelta%3D%28-1%29%5E2-4%5Ccdot1%5Ccdot%28-102%29%5C%5C%20%5CDelta%3D1%2B408%5C%5C%20%5CDelta%3D409%5C%5C%20%5Csqrt%7B%5CDelta%7D%3D%5Csqrt%7B409%7D%5C%5C%5C%5C%20x_1%3D%5Cfrac%7B-%28-1%29-%5Csqrt%7B409%7D%7D%7B2%7D%5C%5C%20x_1%3D%5Cfrac%7B1-%5Csqrt%7B409%7D%7D%7B2%7D%5C%5C%5C%5C%20x_2%3D%5Cfrac%7B-%28-1%29%2B%5Csqrt%7B409%7D%7D%7B2%7D%5C%5C%20x_2%3D%5Cfrac%7B1%2B%5Csqrt%7B409%7D%7D%7B2%7D%5C%5C%5C%5C%20" />0 \ \ x ^ 2 + x - 2x - 2>0 \ \ x(x + 1) - 2(x + 1)>0 \ \ (x - 2)(x + 1)>0 \ \ x \ in( - \ infty, - 1) \ cup(2, \ infty) \ \ 10 ^ 2 = x ^ 2 - x - 2 \ \ 100 = x ^ 2 - x - 2 \ \ x ^ 2 - x - 102 = 0 \ \ \ Delta = ( - 1) ^ 2 - 4 \ cdot1 \ cdot( - 102) \ \ \ Delta = 1 + 408 \ \ \ Delta = 409 \ \ \ sqrt{ \ Delta} = \ sqrt{409} \ \ \ \ x_1 = \ frac{ - ( - 1) - \ sqrt{409}}{2} \ \ x_1 = \ frac{1 - \ sqrt{409}}{2} \ \ \ \ x_2 = \ frac{ - ( - 1) + \ sqrt{409}}{2} \ \ x_2 = \ frac{1 + \ sqrt{409}}{2} \ \ \ \ " alt = " \ \ \ log(x ^ 2 - x - 2) = 2 \ \ x ^ 2 - x - 2>0 \ \ x ^ 2 + x - 2x - 2>0 \ \ x(x + 1) - 2(x + 1)>0 \ \ (x - 2)(x + 1)>0 \ \ x \ in( - \ infty, - 1) \ cup(2, \ infty) \ \ 10 ^ 2 = x ^ 2 - x - 2 \ \ 100 = x ^ 2 - x - 2 \ \ x ^ 2 - x - 102 = 0 \ \ \ Delta = ( - 1) ^ 2 - 4 \ cdot1 \ cdot( - 102) \ \ \ Delta = 1 + 408 \ \ \ Delta = 409 \ \ \ sqrt{ \ Delta} = \ sqrt{409} \ \ \ \ x_1 = \ frac{ - ( - 1) - \ sqrt{409}}{2} \ \ x_1 = \ frac{1 - \ sqrt{409}}{2} \ \ \ \ x_2 = \ frac{ - ( - 1) + \ sqrt{409}}{2} \ \ x_2 = \ frac{1 + \ sqrt{409}}{2} \ \ \ \ " align = "absmiddle" class = "latex - formula">.
1. <img src="https://tex.z-dn.net/?
Hola 2. Logx - 2log(x - 1) = 0 dividimos en 2 2. (logx) / 2 - 2(log(x - 1)) / 2 = 0 / 2 log(x) - log(x - 1) = 0 sumamos log(x - 1) log(x) - log(x - 1) + log(x - 1) = 0 + log(x - 1) Al iliminar terminos iguales con…
La función Log está definida sólo para valores >0, por lo que : x - 2>0 ; x>2 El dominio es : para todo x >2, o : el intervalo abierto de x : )2 ; + infinito( PD : Para tus detalles adicionales : a) La respuesta b) es…
0 \ \ \ log x - \ log5 = 1 \ \ \ log \ frac{x}{5} = 1 \ \ 10 = \ frac{x}{5} \ \ x = 50 \ \ " alt = " \ \ \ log x = \ log5 + 1 \ \ x>0 \ \ \ log x - \ log5 = 1 \ \ \ log \ frac{x}{5} = 1 \ \ 10 = \ frac{x}{5} \ \ x = 50…
Sea la expresión : (Logx)² + Logx - 2 = 0 Hacemos cambio de variable Logx = t t² + t - 2 = 0 - - - >Factorizando por el metodo del aspa t 2 = > 2t X t - 1 = > - t - - - - - - t - - - - > Los factores son : (t + 2)(t -…