How do you solve \[{{2}^{x}}\cdot 5={{10}^{x}}\]?

Hint: This question is from the topic of pre-calculus. In solving this question, we will first take log to the both sides of the equation. After that, we will use a formula of logarithm that is \[\log \left( a\times b \right)=\log a+\log b\] to separate the terms. After that, we will use the formula of logarithm that is \[\log {{a}^{n}}=n\log a\] to remove the x from the power. After using the formula \[\log \left( a\times b \right)=\log a+\log b\], we will solve the equation and find the value of x.

Complete step by step solution:
Let us solve this question.
In this question, we have to solve the equation \[{{2}^{x}}\cdot 5={{10}^{x}}\]. Or, we can say we have to find the value of x from the given equation.
The given equation is
\[{{2}^{x}}\cdot 5={{10}^{x}}\]
Now, taking ‘log’ to the both side of equation, we can write the above equation as
\[\Rightarrow \log \left( {{2}^{x}}\cdot 5 \right)=\log {{10}^{x}}\]
Now, using the formula of logarithms that is \[\log \left( a\times b \right)=\log a+\log b\], we can write the above equation as
\[\Rightarrow \log {{2}^{x}}+\log 5=\log {{10}^{x}}\]
Now, using the formula of logarithm that is \[\log {{a}^{n}}=n\log a\], we can write the above equation as
\[\Rightarrow x\log 2+\log 5=x\log 10\]
Now, taking the terms of x to the right side of equation, we get
\[\Rightarrow \log 5=x\log 10-x\log 2\]
\[\Rightarrow \log 5=x\left( \log 10-\log 2 \right)\]
Using the formula \[\log \left( \dfrac{a}{b} \right)=\log a-\log b\], we can write the above equation as
\[\Rightarrow \log 5=x\left( \log \dfrac{10}{2} \right)\]
The above equation can also be written as
\[\Rightarrow \log 5=x\left( \log 5 \right)\]
Now, dividing \[\log 5\] to the both side of the equation, we get
\[\Rightarrow 1=x\]
Or, we can say that
\[\Rightarrow x=1\]
Hence, we have the equation \[{{2}^{x}}\cdot 5={{10}^{x}}\] and found the value of x as 1.

Note: We should have a better knowledge in logarithmic functions so we can solve this type of question easily. The following formulas should be kept remembered to solve this type of question:
\[\log \left( a\times b \right)=\log a+\log b\]
\[\log \left( \dfrac{a}{b} \right)=\log a-\log b\]
\[\log {{a}^{n}}=n\log a\]
We can solve this by an alternate method.
The equation which we have to solve is
\[{{2}^{x}}\cdot 5={{10}^{x}}\]
The above equation can also be written as
\[\Rightarrow 5=\dfrac{{{10}^{x}}}{{{2}^{x}}}\]
The above equation can also be written as
\[\Rightarrow 5={{\left( \dfrac{10}{2} \right)}^{x}}\]
\[\Rightarrow 5={{\left( 5 \right)}^{x}}\]
As we know that 5 can also be written as 5 to the power 1, so we can write the above equation as
\[\Rightarrow {{5}^{1}}={{\left( 5 \right)}^{x}}\]
We can see that the base (that is 5) is the same on both sides of the equation, so we can say powers will also be the same. So, we can write
\[\Rightarrow 1=x\]
Or, we can write
 \[\Rightarrow x=1\]
Hence, we can use this method too because we have got the same answer from this method.