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Question

Question: How do I solve \(5{{e}^{2x}}=500\)?...

How do I solve 5e2x=5005{{e}^{2x}}=500?

Explanation

Solution

Here we need to solve for x. We simplify the given equation using the definition of logarithm and exponential. After that we apply the logarithm power rule mlogay=loga(ym)m{\log _a}y = {\log _a}\left( {{y^m}} \right) to get the desired result.

Formula used: mlogay=loga(ym)m{{\log }_{a}}y={{\log }_{a}}\left( {{y}^{m}} \right)

Complete step by step solution:
The given equation says that 5e2x=5005{{e}^{2x}}=500
We divide it by 5 and write it as e2x=100{{e}^{2x}}=100 … (i)
The given expression is a function of x in which a constant number is raised to the variable x. This time of function is called an exponential function.
The general form of an exponential function with variable x is given as f(x)=axf(x)={{a}^{x}}, where f(x) is a function of x and a is the constant. The variable x is called exponent and the constant ‘a‘ is referred to as the base of the exponent.
To solve the given equation we will logarithmic function.
Logarithmic function is the inverse of exponential function.
For exponential function y=axy={{a}^{x}}, the corresponding logarithmic function is x=logayx={{\log }_{a}}y.
In this case, a=ea=e and y=100y=100.
Then we can write that 2x=loge1002x={{\log }_{e}}100
x=12loge100\Rightarrow x=\dfrac{1}{2}{{\log }_{e}}100
By using the property of logarithmic functions we get that x=12loge100=loge(10012)x=\dfrac{1}{2}{{\log }_{e}}100={{\log }_{e}}\left( {{100}^{\dfrac{1}{2}}} \right)
On simplifying further we get x=loge(10)x={{\log }_{e}}\left( 10 \right)

Therefore, the solution for the given equation is x=loge(10)x={{\log }_{e}}\left( 10 \right).

Note: We can also write the solution for the question as follows.
One of the properties of logarithmic functions say that loge(ex)=x{{\log }_{e}}\left( {{e}^{x}} \right)=x.
Therefore, if we apply the logarithmic function to equation (i), then we get loge(e2x)=loge100{{\log }_{e}}\left( {{e}^{2x}} \right)={{\log }_{e}}100 2x=loge100\Rightarrow 2x={{\log }_{e}}100 (by using the above property).