How do you solve $2 {log}_{b} 4 + {log}_{b} 5 - {log}_{b} 10 = {log}_{b} x$ $2log_b4+log_b5-log_b10=log_bx$ ?

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Answer 1 · 2016-07-17T16:35:34

x=8

Since $n log a = {log a}^{n}$ $nloga=loga^n$ you have:

${log}_{b} 4^{2} + {log}_{b} 5 - {log}_{b} 10 = {log}_{b} x$ $log_b 4^2+log_b 5-log_b 10=log_b x$ .

Since $log p + log q = log (p q)$ $logp+logq=log(pq)$ you have:

${log}_{b} (16 \cdot 5) - {log}_{b} 10 = {log}_{b} x$ $log_b (16*5)-log_b 10=log_b x$ .

Since $log p - log q = log (\frac{p}{q})$ $logp-logq=log(p/q)$ you have:

${log}_{b} (\frac{80}{10}) = {log}_{b} x$ $log_b(80/10)=log_b x$ .

Then ${log}_{b} 8 = {log}_{b} x$ $log_b8=log_b x$

so x=8

How do you solve 2log_b4+log_b5-log_b10=log_bx2logb4+logb5−logb10=logbx2log_b4+log_b5-log_b10=log_bx?