How do you solve ${log}_{5} 2 + 2 {log}_{5} t = {log}_{5} (3 - t)$ $log_5 2 + 2 log_5t= log_5(3-t)$ ?

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Answer 1 · 2016-05-04T17:04:54

$t = - \frac{3}{2} or t = 1$ $t=-3/2 or t=1$

Use Property : ${log}_{b} (x y) = {log}_{b} x + {log}_{b} y$ $log_b(xy)=log_bx+log_by$ and ${log}_{b} x^{n} = n {log}_{b} x$ $log_bx^n=nlog_bx$

${log}_{5} 2 + 2 {log}_{5} t = {log}_{5} (3 - t)$ $log_5 2 +2log_5 t =log_5(3-t)$

${log}_{5} 2 + {log}_{5} t^{2} = {log}_{5} (3 - t)$ $log_5 2+log_5 t^2 = log_5(3-t)$

${log}_{5} 2 t^{2} = {log}_{5} (3 - t)$ $log_5 2t^2= log_5(3-t)$

$2 t^{2} = 3 - t$ $2t^2 = 3-t$

$2 t^{2} + t - 3 = 0$ $2t^2+t-3=0$

$(2 t + 3) (t - 1) = 0$ $(2t+3)(t-1)=0$

$t = - \frac{3}{2} or t = 1$ $t=-3/2 or t=1$

How do you solve log52+2log5t=log5(3−t)log_5 2 + 2 log_5t= log_5(3-t)?