How do you solve ${log}_{b} 30 - {log}_{b} 5^{2} = {log}_{b} x$ $log_b30-log_b5^2=log_bx$ ?

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Answer 1 · 2016-07-25T21:22:57

$x = \frac{6}{5}$ $x=6/5$

${log}_{b} 30 - {log}_{b} 5^{2} = {log}_{b} x ?$ $log_b 30-log_b 5^2=log_b x ?$

$so ; {log}_{a} (\frac{b}{c}) = {log}_{a} b - {log}_{a} c$ $"so ;" log _a( b/c)=log_a b-log_a c$

${log}_{b} (\frac{30}{25}) = {log}_{b} x$ $log _b (30/25)=log _b x$

$\frac{30}{25} = x$ $30/25=x$

$x = \frac{6}{5}$ $x=6/5$

How do you solve log_b30-log_b5^2=log_bxlogb30−logb52=logbxlog_b30-log_b5^2=log_bx?