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Question

Question: How do you solve \( {5^x} = 17 \) ?...

How do you solve 5x=17{5^x} = 17 ?

Explanation

Solution

Hint : In order to determine the value of the above question ,first take logarithm on both the sides of equation and use logmn=nlogm\log {m^n} = n\log m and then put the value of log17andlog5\log 17\,and\log 5 using calculator to get your desired answer.
FORMULAE:
nlogm=logmn   n\log m = \log {m^n} \;
logb(mn)=logb(m)logb(n){\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)

Complete step-by-step answer :
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
First, we are going to rewrite the number
5x=17{5^x} = 17
Now taking logarithm on both the sides
log(5x)=log(17)\log \left( {{5^x}} \right) = \log \left( {17} \right)
Using proper of logarithm logmn=nlogm\log {m^n} = n\log m
So,

log(5x)=log(17) x=log(17)log5   \log \left( {{5^x}} \right) = \log \left( {17} \right) \\\ x = \dfrac{{\log \left( {17} \right)}}{{\log 5}} \;

Now using calculator calculating the value of log17andlog5\log 17\,and\log 5 which comes to be
log17=1.23044892138 log5=0.69897000433   \log 17 = 1.23044892138 \\\ \log 5 = 0.69897000433 \;
Putting values in equation (1)

x=1.230448921380.69897000433 x=1.7603744273   \Rightarrow x = \dfrac{{1.23044892138}}{{0.69897000433}} \\\ \Rightarrow x = 1.7603744273 \;

Therefore, the solution is x=1.7603744273x = 1.7603744273 .
So, the correct answer is “ x=1.7603744273x = 1.7603744273 ”.

Note : 1. Value of the constant ”e” is equal to 2.71828.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
logb(mn)=logb(m)+logb(n){\log _b}(mn) = {\log _b}(m) + {\log _b}(n)
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
logb(mn)=logb(m)logb(n){\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
nlogm=logmnn\log m = \log {m^n}