Question

How do you solve \[{{(0.3)}^{1+x}}={{1.7}^{2x-1}}\] ?

Answer 1

Hint: We are given an expression which we have to solve for \[x\]. We will be using the logarithm function first. Then, the property of the logarithm function \[\log {{a}^{m}}=m\log a\] will be used. We will simplify the expression and rewrite the expression in terms of \[x\]. Then putting the log values in the expression, we will have the value of \[x\].

Complete step by step solution:
According to the given question, we are asked to solve the given expression for \[x\].
The expression we have is,
\[{{(0.3)}^{1+x}}={{1.7}^{2x-1}}\]------(1)
Now, applying logarithm function on either side of the equation (1), we get,
\[\log {{(0.3)}^{1+x}}=\log {{1.7}^{2x-1}}\]------(2)
We know that, \[\log {{a}^{m}}=m\log a\], using this in the above equation, we get,
\[\Rightarrow (1+x)\log 0.3=(2x-1)\log 1.7\]-----(3)
Now, we will open the brackets on either side in the equation (3), we have,
\[\Rightarrow \log 0.3+x\log 0.3=2x\log 1.7-\log 1.7\]
We now have to separate the terms with \[x\] and the constant terms and then we will write the expression in terms of \[x\]. Adding \[\log 1.7\] on both the sides, we get the new expression as,
\[\Rightarrow \log 0.3+x\log 0.3+\log 1.7=2x\log 1.7-\log 1.7+\log 1.7\]-----(4)
In RHS, \[\log 1.7\] will get cancelled and in the LHS, \[\log 1.7\] will get added, we have,
\[\Rightarrow \log 0.3+x\log 0.3+\log 1.7=2x\log 1.7\]----(5)
Now, subtracting the equation by \[x\log 0.3\] on either side of the equal to sign, we get,
\[\Rightarrow \log 0.3+x\log 0.3+\log 1.7-x\log 0.3=2x\log 1.7-x\log 0.3\]
Solving further, we get the expression as,
\[\Rightarrow \log 0.3+\log 1.7=2x\log 1.7-x\log 0.3\]----(6)
We will now write the equation (6) in terms of \[x\] , for that we will be taking \[x\] common from RHS in the equation (6), we have,
\[\Rightarrow \log 0.3+\log 1.7=x(2\log 1.7-\log 0.3)\]
Writing the expression in terms of \[x\], we get,
\[\Rightarrow x=\dfrac{\log 0.3+\log 1.7}{2\log 1.7-\log 0.3}\]-----(7)
We will put the values of these logarithm terms and we know that,
\[\log 0.3=-0.522\] and \[\log 1.7=0.230\]
We get,
\[\Rightarrow x=\dfrac{-0.522+0.230}{2(0.230)-(-0.522)}\]
Solving the expression above, we get the value of \[x\] as,
\[\Rightarrow x=\dfrac{-0.292}{2(0.230)+(0.522)}\]
\[\Rightarrow x=\dfrac{-0.292}{0.980}=-0.297\]

Therefore, the value of \[x=-0.297\].

Note: The logarithm function should be carefully and step wise applied to the given expression. Also, the value of logarithm terms should be correctly substituted in the expression as well as the calculation should also be done correctly.