1. [tex]\begin{gathered} {\log _3}\left( {2x + 5} \right) = 2 \hfill \\ \Rightarrow \frac{{\log \left( {2x + 5} \right)}}{{\log 3}} = 2 \hfill \\ \Rightarrow \log \left( {2x + 5} \right) = 2\log 3 = \log {3^2} \hfill \\ \Rightarrow \log \left( {2x + 5} \right) = \log 9 \hfill \\ \Rightarrow 2x + 5 = 9 \hfill \\ \Rightarrow 2x = 4 \hfill \\ \Rightarrow x = 2 \hfill \\ \end{gathered} [/tex]
3. [tex]\[\begin{gathered} \log \left[ {\left( {x - 1} \right)\left( {x + 5} \right)} \right] = 2 \hfill \\ \Rightarrow \log \left[ {\left( {x - 1} \right)\left( {x + 5} \right)} \right] = \log 100 \hfill \\ \Rightarrow \left( {x - 1} \right)\left( {x + 5} \right) = 100 \hfill \\ \Rightarrow {x^2} + 4x - 5 = 100 \hfill \\ \Rightarrow {x^2} + 4x - 105 = 0 \hfill \\ \Rightarrow x = - 2 + \sqrt {109} \vee x = - 2 - \sqrt {109} \hfill \\ \end{gathered} [/tex]
las otras se hacen igual...
Respuesta:
yo lo hice así
log 3 (2x+5)=2
3²=2x+5
9=2x+5
9-5=2x
4=2x
x=2
Explicación paso a paso:
me salio igual que el de arriba
