Question #53728

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Answer 1 · 2018-02-10T20:28:30

#x=(1-log_10(3))/log_10(1.32)# As an exact value

#x~~4.337# to 3 decimal places as an approximate value

Before we start note that #log(axxb)=log(a)+log(b)#
Also that #log(a^b)=blog(a)#

Simplify a bit first by dividing both sides by 100 giving:

#50=15(1.32)^x#

Simplify further by dividing both sides by 5

#10=3(1.32)^x#

Take logs of both sides. I chose logs to base #10->log_10#

This is because #log_10(10)=1#

#log_10(10)=log_10(3)+log_10(1.32^x)#

#1=log_10(3)+xlog_10(1.32)#

Subtract #log_10(3)# from both sides

#1-log_10(3)=xlog_10(1.32)#

#x=(1-log_10(3))/log_10(1.32)# As an exact value

#x~~4.336582.....#

#x~~4.337# to 3 decimal places As an approximate value