How do you solve # 3^(2x+1) = 5#?

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Answer 1 · 2016-08-08T09:07:56

#x=0.2326#.

#3^(2x+1)=5#

#:. 3^(2x)*3=5#

#:. 3^(2x)=5/3#

#:. log_10 3^(2x)=log_10 (5/3)#

#:.2x*log_10 3=log_10 5-log_10 3#

#:. 2x(0.4771)=0.6990-0.4771#................[Using, Log Tables].

#:. 2x(0.4771)=0.2219#

#:. log_10{(2x)(0.4771)}=log_10 0.2219#

#:. log_10 2+log_10 x+log_10 0.4771=log_10 0.2219#

#:. 0.3010+log_10x+(bar1).6786=(bar1).3461.#

#:.log_10x+(bar1).9796=(bar1).3461#

#:. log_10x=0.3461-0.9796=-0.6335=(bar1).3665#

#:. x#=Anti-#log_10 (bar1).3665#

#:. x=0.2326#.

Answer 2 · 2016-08-11T21:56:49

#x = 0.232487#

We have an exponential equation, but 5 is not one of the powers of 3, so logs are indicated, especially if the variable is in the index.

#3^(2x+1) = 5#

#log 3^(2x+1) = log 5#

#(2x+1)log3 =log 5#

#2x+1 = log5/log#

There is no law for simplifying #(log)/log), use a calculator or tables.

#2x +1 = 1.4649735#

#2x = 0.464974#

#x = 0.232487#