Respuesta :
well, assuming is not inches, and instead is ln(x).
[tex]\bf \textit{Logarithm Cancellation Rules}\\\\ log_a a^x= x\qquad \qquad \boxed{a^{log_ax}=x}\\\\ -------------------------------\\\\ 6+ln(x)=8\implies ln(x)=2\implies log_e(x)=2 \\\\\\ e^{log_e(x)}=e^2\implies x=e^2[/tex]
[tex]\bf \textit{Logarithm Cancellation Rules}\\\\ log_a a^x= x\qquad \qquad \boxed{a^{log_ax}=x}\\\\ -------------------------------\\\\ 6+ln(x)=8\implies ln(x)=2\implies log_e(x)=2 \\\\\\ e^{log_e(x)}=e^2\implies x=e^2[/tex]
Answer:
[tex]x=7.39[/tex]
Step-by-step explanation:
we are given equation as
[tex]6+ln(x)=8[/tex]
Since, we have to solve for x
so, we will isolate x on anyone side
Firstly, we will subtract both sides by 6
[tex]6+ln(x)-6=8-6[/tex]
[tex]ln(x)=2[/tex]
now, we can take exponent of e on both sides
[tex]e^{ln(x)}=e^2[/tex]
now, we can use property of log
and we get
[tex]e^{ln(a)}=a[/tex]
we can use it
[tex]x=e^2[/tex]
So, solution is
[tex]x=7.39[/tex]
