Question 1180949
<br>
The function you show is not an exponential function:<br>
f(x) = 5(1/3)^2x-1 is {{{f(x) = 5(1/3)^2x-1}}}<br>
It's astonishing to me how few students who post questions about this level of math seem to realize that the proper use of parentheses is important...!<br>
Nevertheless, there is some good algebra to be learned here; so I will go ahead and work with the function you INTENDED to show.<br>
f(x) = 5(1/3)^(2x-1) = {{{f(x) = 5(1/3)^(2x-1)}}}<br>
We are to write this as an equivalent function in the form {{{a(d)^x}}}<br>
To do that, we need to get rid of the "2" and the "-1" in the exponent.<br>
To get rid of the "-1" in the exponent, multiply AND divide the expression by (1/3):<br>
{{{5/(1/3)=5*3=15}}}
{{{((1/3)^(2x-1))(1/3)=(1/3)^(2x)}}}<br>
The expression is now<br>
{{{15(1/3)^(2x)}}}<br>
To make the exponent x instead of 2x, use the power-to-a-power rule:<br>
{{{15(1/3)^(2x)=15((1/3)^2)^x=15(1/9)^x}}}<br>
ANSWER: {{{f(x) = 5(1/3)^(2x-1)=15(1/9)^x}}}<br>