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Question

Question: How do you solve the algebraic expression \({5^{2x}} = 8\)?...

How do you solve the algebraic expression 52x=8{5^{2x}} = 8?

Explanation

Solution

This problem deals with finding the solution of xx, with the help and applications of logarithms to the given exponential equation. Here some basic and fundamental identities or properties of logarithms are used in order to solve the given exponential equation such as:
logan=nloga\Rightarrow \log {a^n} = n\log a

Complete step-by-step solution:
Given an expression in exponential form where the exponent is varying in the variable xx.
Consider the given exponential equation below as shown:
52x=8\Rightarrow {5^{2x}} = 8
Now applying natural logarithms on both sides of the given above equation, as shown below:
loge52x=loge8\Rightarrow {\log _e}{5^{2x}} = {\log _e}8
Here we know that the basic identity or the property of logarithms which is logan=nloga\log {a^n} = n\log a, applying this property to the left hand side of the above equation, as shown below:
2xloge5=loge8\Rightarrow 2x{\log _e}5 = {\log _e}8
Now divide the above equation by loge5{\log _e}5, on both sides of the above equation, as shown below:
2x=loge8loge5\Rightarrow 2x = \dfrac{{{{\log }_e}8}}{{{{\log }_e}5}}
Now simplifying the right hand side of the above equation, by substituting the values of loge8=2.079{\log _e}8 = 2.079 and loge5=1.609{\log _e}5 = 1.609, in the above equation as shown below:
2x=2.0791.609\Rightarrow 2x = \dfrac{{2.079}}{{1.609}}
Simplifying the above expression on the right hand side of the equation and then dividing the equation by 2, to get the value of xx, as given below:
2x=1.292\Rightarrow 2x = 1.292
x=0.646\therefore x = 0.646
The solution of 52x=8{5^{2x}} = 8 is equal to 0.646.

Note: Please note that the above problem is solved with the help of logarithms, that is by applying the logarithms on both sides of the equation to get the value of the variable xx. Likewise there are other similar properties of logarithms which can be used as applications in order to solve such kind of problems such as:
logab=loga+logb\Rightarrow \log ab = \log a + \log b
log(ab)=logalogb\Rightarrow \log \left( {\dfrac{a}{b}} \right) = \log a - \log b
elogea=a\Rightarrow {e^{{{\log }_e}a}} = a