How do you solve #1/2=10^(1.5x)#?

2 Answers
Srinivas
Mar 13, 2016

x = -0.200686663

Explanation:

#1 / 2 = 10^(1.5 * x)#
Take Log on both sides
#log (1 / 2) = 1.5 x * log 10#
(Hope you are aware of the relationship
#log a^x = x * log a#)
Now #log ( 1 / 2) = log 2^-1 = -1 * log 2#
log 2 = 0.3010300 (this you can get from any log tables or calculator)
So, #-0.3010300 = 1.5 x#
or #x = -0.3010300 / 1.5 = -0.200686663#
Hope it is clear

P dilip_k
Mar 13, 2016

#:.x=-2/3log_10 2#

Explanation:

tking #log_10# on both sides

#log_10(1/2)=log_10 10^(1.5x)#

#=>log_10 1-log_10 2=log_10 10^(1.5x)#

#=>0-log_10 2=x*1.5xxlog_10 10#

#=>0-log_10 2=(3x)/2#
#:.x=-2/3log_10 2#

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