Question

Solve for x to five decimal places if 5^(2x-1)=357 Substitute your x value answer back into original and show that it will make each side equal.

Answer 1

that the base e log if both sides

\[\ln(5^{2x -1} = (2x -1)\ln(5)\]

using this fact then

\[(2x -1)\ln(5) = \ln (357)\]

\[(2x -1) = \ln(357)/\ln(5)\]

the right hand side is a number then

\[2x = 1+\ln(357)/\ln(5)\]

that should make it easy for you to find the value of x

Answer 2

We have
\[5^{(2x-1)}=357\]
Take log on both sides, we get
\[\log_{10}(5^{(2x-1)})=\log_{10}(357)\]
We know that
\[\log_a b^c=c\times \log_a b\]
so using this property here. We get
\[(2x-1) \log_{10} 5= \log_{10} 357\]
We can use either log tables or a calculator to find the values of \(\log_{10} 5,\log_{10} 357\)
\[\log_{10} 5=0.6989\]
and
\[\log_{10} 357=2.55266\]
so we have
\[(2x-1) \times 0.6989=2.55266\]
we get
\[x=2.32602\]

Answer 3

alll the above methods are correct, but if you do more than two or three of these you should remember the change of base formula which tells you how to solve for a variable that is in the exponent:
\[b^x=A\implies x=\frac{\ln(A)}{\ln(b)}\] in english,
"the log of the total divided by the log of the base"

Answer 4

so if you see another similar problem to
\[5^{2x-1}=357\] go right to
\[2x-1=\frac{\ln(357)}{\ln(5)}\]