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7^2x-1= 316. exponential equations...
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\[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]
\[7^{2x-1}=316\iff 2x-1=\frac{\ln(316)}{\ln(7)}\]
use a calculator to compute the right hand side, then solve for \(x\) by adding one and dividing the result by 2
let me know what you get as a final answer
i got 1.48
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then there is a mistake somewhere
i did what you said to do
here is \[\frac{\ln(316)}{\ln(7)}\] http://www.wolframalpha.com/input/?i=ln%28316%29%2Fln%287%29 is the what you got ?
no
about 2.95866 add 1 to get 3.95866 then divide by 2
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perhaps you put it in the calculator wrong you have to close the parentheses after the log on top
i got 1.979
that is it
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