Logarithm help
- Thread starter 8dltndus
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Did you notice that 49 = 72? What happens if you use the change-of-base formula on the right-hand side, changing it into a base-7 expression?
HallsofIvy
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Yet another way of looking at this:
If \(\displaystyle x= log_7(y- 6)\) then \(\displaystyle y- 6= 7^x\) or \(\displaystyle y= 7^x+ 6\).
If also \(\displaystyle x= log_{49}(2y)\) then \(\displaystyle 2y= 49^x\) or \(\displaystyle y= \frac{1}{2}49^x\).
But, as has been pointed out, already, \(\displaystyle 49= 7^2\) so \(\displaystyle 49^x= (7^2)^x= 7^{2x}= (7^x)^2\) so \(\displaystyle y= \frac{1}{2}(7^x)^2\).
So we have \(\displaystyle y= 7^x+ 6= \frac{1}{2}(7^x)^2\).
Let \(\displaystyle z= 7^x\) and we have the quadratic equation \(\displaystyle z+ 6= \frac{1}{2}z^2\) which is the same as \(\displaystyle z^2- 2z- 12= 0\).
Solve that quadratic equation for z, then solve \(\displaystyle 7^x= z\) for x.
If \(\displaystyle x= log_7(y- 6)\) then \(\displaystyle y- 6= 7^x\) or \(\displaystyle y= 7^x+ 6\).
If also \(\displaystyle x= log_{49}(2y)\) then \(\displaystyle 2y= 49^x\) or \(\displaystyle y= \frac{1}{2}49^x\).
But, as has been pointed out, already, \(\displaystyle 49= 7^2\) so \(\displaystyle 49^x= (7^2)^x= 7^{2x}= (7^x)^2\) so \(\displaystyle y= \frac{1}{2}(7^x)^2\).
So we have \(\displaystyle y= 7^x+ 6= \frac{1}{2}(7^x)^2\).
Let \(\displaystyle z= 7^x\) and we have the quadratic equation \(\displaystyle z+ 6= \frac{1}{2}z^2\) which is the same as \(\displaystyle z^2- 2z- 12= 0\).
Solve that quadratic equation for z, then solve \(\displaystyle 7^x= z\) for x.