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How do you solve #log_5(x) + log_3(x) =1#?

1 Answer

Dec 4, 2015

I found: #x=1.921148#

Explanation:

We can change base:
#(ln(x))/(ln(5))+(ln(x))/(ln(3))=1#
#ln(x)(1/ln(5)+1/(ln(3)))=1#
#ln(x)[1.531574]=1#
#ln(x)=1/1.531574=0.652923#

use the definition of log:

#x=e^0.652923=1.921148#

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