Evaluate #log_5(10)xxlog_5(x)=log_5(100)#?

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Answer 1 · 2017-03-29T03:25:48

#x=25#

We have:

#log_5(10)xxlog_5(x)=log_5(100)#

#log_5x=log_5(100)/log_5(10)=((log100)/log5)/(log10/log5)=log100/log5xxlog5/log10=log100/log10=2/1=2#

We can now rewrite the log as an exponential, following the form:

#log_a(b)=c=>a^c=b#

#log_5(x)=2=>5^2=x=25#

Answer 2 · 2017-03-29T04:54:18

#x = 25#

#log_5 10 * log_5 x = log_5 100#

# log_5 x = log_5 100/log_5 10#

# log_5 x = log_5 100/log_5 10 = (log_10 100/log_10 5) / (log_10 10/log_10 5)# -> change it base from 5 to 10.

# log_5 x = log_10 100 / log_10 10 = 2#

# log_5 x= 2#

#x = 5^2 =25#