How do you solve \[3\log x = 1.5\]?

Hint: Here the question given is \[3\log x = 1.5\] and we need to solve for it.
First we will simplify it using the properties of logarithm and we will find two different expressions using the properties of logarithm and then use it to one another or substitute it in order to find the value of $x$.

Complete step by step answer:
In this question we are asked to solve the given expression \[3\log x = 1.5\]
Here we can transfer the given number $3$ to the power of $x$ and hold it
Transferring the real number as a power of the variable,
\[ \Rightarrow 3\log (x) = 1.5\]
On rewriting the term and we get,
$ \Rightarrow 3\log (x) = \log ({x^3})$
Then, we get
$ \Rightarrow \log ({x^3}) = 1.5$
Now back to the given expression. We know that any logarithm as a power of number is itself.
That is, ${10^{log(x)}} = x$
With this property, we can interpret or change the expression or also write it as,
$ \Rightarrow {10^{log({x^3})}} = {x^3}$
Now, we got $\log ({x^3}) = 1.5$ and ${10^{log({x^3})}} = {x^3}$
Therefore we can substitute the former in the place later. That is we will put $\log ({x^3}) = 1.5$in the second one.
Hence it becomes,
$ \Rightarrow {x^3} = {10^{1.5}}$
We can expand the RHS by
$ \Rightarrow {10^{0.5}} \times {10^{0.5}} \times {10^{0.5}}$
Also we can expand $x$ too, so we get
$ \Rightarrow x \times x \times x = {10^{0.5}} \times {10^{0.5}} \times {10^{0.5}}$
It ultimately means that,
$ \Rightarrow x = {10^{0.5}}$
Which also can be written as ${10^{\dfrac{1}{2}}}$,
$ \Rightarrow {10^{\dfrac{1}{2}}}$
On rewriting the term and we get
$ \Rightarrow \sqrt {10} $

Therefore the value of $x$ in \[3\log x = 1.5\] is $\sqrt {10} $.

Note: Logarithm is the exponent or power to which a base must be raised to yield a given number. That is, we can say that \[x\] is the logarithm of \[n\] to the base \[b\] if ${b^x} = n$, we can write it as \[x\; = {\text{ }}{\log _b}\;n\] .
For example, ${3^2} = 9$ therefore, $2$ is the logarithm of $9$ to base $3$ , or\[{\text{2 }} = {\text{ }}{\log _3}\;9\]. Like this, since \[{10^2}\; = {\text{ }}100\], then \[2{\text{ }} = {\text{ }}lo{g_{10}}\;100\] .