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Find a so that the graph of $f(x) = \log_a x$ contains the point $(26, 2)$. a = \log_{\sqrt{26}} 2 (Simplify your answer. Type an exact answer, using radicals as needed. Use integers c

          Find a so that the graph of $f(x) = \log_a x$ contains the point $(26, 2)$. 
a = \log_{\sqrt{26}} 2
(Simplify your answer. Type an exact answer, using radicals as needed. Use integers c
        
Find a so that the graph of f(x) =  x contains the point (26, 2). 
a = log√(26) 2
(Simplify your answer. Type an exact answer, using radicals as needed. Use integers c

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Find a so that the graph of f(x)=log_(a)x contains the point (26,2). a= (Simplify your answer. Type an exact answer, using radicals as needed. Use integers Find a so that the graph of f(x)=logx contains the point (26,2) KK a= 1092/26 (Simplify your answer.Type an exact answer, using radicals as needed. Use integers
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Transcript

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00:01 According to gain question, we have 11 x square plus 10 root 3 xy plus y square is equal to 32.
00:10 So we can write here as here we have a is equal to 11, b is equal to 10 root 3 and c is equal to 1.
00:20 So b squared minus 4 ac, this is coming as 10 root 3 square minus 4 multiplied with 11 multiplied with 1.
00:30 This is 100 multiplied with 3 is 300 minus 44 which is greater than 0.
00:35 So this is hyperbola.
00:38 So the curve we will get will be hyperbola.
00:41 So now quat 2 theta, where theta is the rotation angle, a minus c divided by b, so this is 11 minus 1 divided by 10 root 3...
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