Question

por favor son problemas de logaritmo ayudenme

Answer 1

Explicación:

1.

$log_{ \frac{1}{3} }(27)$

$log_{ {3}^{ - 1} }( {3}^{3} )$

$\frac{3}{ - 1}$

$- 3$

2.

$log_{ \sqrt[5]{5} }( \sqrt[6]{25} )$

$log_{ {5}^{ \frac{1}{5} } }( {5}^{ \frac{1}{3} } )$

$\frac{ \frac{1}{3} }{ \frac{1}{5} }$

$\frac{5}{3}$

3.

$e = log_{2}(128) - 3 log_{4}(256) + 2 log_{ \sqrt{3} }(27)$

$e =7 - 3 \times 4 + 2 log_{ {3}^{ \frac{1}{2} } }( {3}^{3} )$

$e = 7 - 12 + log_{ {3}^{ \frac{1}{2} } }( {3}^{6} )$

$e = - 5 + \frac{6}{ \frac{1}{2} }$

$e = - 5 + 12$

$e = 7$

4.

$m = {1024}^{ log_{2}(3) }$

$m = {2}^{10 log_{2}(3) }$

$m = {2}^{ log_{2}( {3}^{10} ) }$

$m = {3}^{10}$

5.

$log_{ \sqrt[3]{2} }(64)$

$log_{ {2}^{ \frac{1}{3} } }( {2}^{6} )$

$\frac{6}{ \frac{1}{3} }$

$18$

6.

$log(x) = 4 - log(625)$

$log_{10}(x) + log_{10}(625) = 4$

$log_{10}(625x) = 4$

${10}^{ log_{10}(625x) } = {10}^{4}$

$625x = 10 \: 000$

$x = 16$

7.

$p = \frac{ log_{1000}(125) }{4 log(25) }$

$p = \frac{ log_{ {10}^{3} }( {5}^{3} ) }{4 log_{10}( {5}^{2} ) }$

$p = \frac{ \frac{3}{3} log_{10}(5) }{4 \times 2 log_{10}(5) }$

$p = \frac{1}{8}$

8.

$a = log_{4}( log_{5}(2 + log_{5}(125) {)}^{2} {)}^{2}$

$a = log_{4}( log_{5}(2 + 3 {)}^{2} ) {)}^{2}$

$a = log_{4}( log_{5}(25) {)}^{2}$

$a = log_{4}(2 {)}^{2}$

$a = 1$

9.

${x}^{ \frac{1}{2} } = log_{2}( log_{2}(256) )$

${x}^{ \frac{1}{2} } = log_{2}(8)$

${x}^{ \frac{1}{2} } = 3$

$x = \sqrt{3}$

10.

${3}^{2 log_{3}(x - 2) } = 16$

${3}^{ log_{3}(x - 2 {)}^{2} } = 16$

$(x - 2 {)}^{2} = 16$

${x}^{2} - 4x + 4 = 16$

${x}^{2} - 4x - 12 = 0$

$x = \frac{ - ( - 4) + \sqrt{( - 4 {)}^{2} - 4(1)( - 12)} }{2(1)}$

$x = \frac{4 + \sqrt{64} }{2}$

$x = \frac{12}{2}$

$x = 6$

11.

${5}^{ log_{5}(4x - 5) } + {2}^{ log_{2}(2x + 1) } = 4(x + 6)$

$4x - 5 + 2x + 1 = 4x + 24$

$2x = 28$

$x = 14$

12.

$p = {2}^{ log_{3} {8}^{ log_{4}(3) } }$

$p = {2}^{ log_{3} {3}^{ log_{4}(8) } }$

$p = {2}^{ log_{3} {3}^{ log_{ {2}^{2} } {2}^{3} } }$

$p = {2}^{ log_{3} {3}^{ \frac{3}{2} } }$

$p = {2}^{ \frac{3}{2} log_{3}(3) }$

$p = {2}^{ \frac{3}{2} }$

$p = \sqrt{2} \times \sqrt{ {2}^{2} }$

$p = 2 \sqrt{2}$

psdt: si te ayudé sígueme y ponme como mejor respuesta :)