Question

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

Answer 1

`x=0.2326`.

Explanation:

`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

`x = 0.232487`

Explanation:

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`