How do you solve ${log}_{5} x + {log}_{10} x = 4$ $log_5 x + log_10 x = 4$ ?

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Answer 1 · 2016-03-25T07:21:09

$x = 44.228$ $x=44.228$

As ${log}_{b} x = \frac{{log}_{a} x}{{log}_{a} b}$ $log_b x=log_ax/log_ab$ , we have

${log}_{5} x + {log}_{10} x = 4$ $log_5x+log_10x=4$ can be written as

$\frac{{log}_{10} x}{{log}_{10} 5} + {log}_{10} x = 4$ $log_10x/log_10 5+log_10x=4$ or

$\frac{{log}_{10} x}{0.699} + {log}_{10} x = 4$ $log_10x/0.699+log_10x=4$ or

${log}_{10} x + 0.699 \times {log}_{10} x = 4 \times 0.699$ $log_10x+0.699xxlog_10x=4xx0.699$ or

${log}_{10} x (1 + 0.699) = 4 \times 0.699$ $log_10x(1+0.699)=4xx0.699$ or

${log}_{10} x = \frac{4 \times 0.699}{1.699} = 1.6457$ $log_10x=(4xx0.699)/1.699=1.6457$

$x = 10^{1.6457} = 44.228$ $x=10^(1.6457)=44.228$

How do you solve log5x+log10x=4log_5 x + log_10 x = 4?