What is the answer to t(n)=4t(n^2)+n^2log_{10}(n) ?

The answer to t(n)=4t(n^2)+n^2log_{10}(n) is t=(nlog_{10}(n))/(1-4n);\quad n\ne 0,n\ne 1/4

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t(n)=4t(n^2)+n^2log_{10}(n)

t (n) = 4 t (n^{2}) + n^{2} lo g_{10} (n)

t = \frac{n lo g _{10} ( n )}{1 - 4 n}; n \neq = 0, n \neq = \frac{1}{4}

Solution steps

t (n) = 4 t (n^{2}) + n^{2} lo g_{10} (n)

t (n) = 4 n^{2} t + n^{2} lo g_{10} (n)

Expand

t (n) : n t

n t = 4 n^{2} t + n^{2} lo g_{10} (n)

Move

4 n^{2} t

to the left side

n t - 4 n^{2} t = n^{2} lo g_{10} (n)

Factor

n t - 4 n^{2} t : n t (1 - 4 n)

n t (1 - 4 n) = n^{2} lo g_{10} (n)

Divide both sides by

n (1 - 4 n); n \neq = 0, n \neq = \frac{1}{4}

t = \frac{n lo g _{10} ( n )}{1 - 4 n}; n \neq = 0, n \neq = \frac{1}{4}

2 y - 3 = 5 + 2 (y - 1)

\frac{3 x}{2} - \frac{5}{3} = x - 1

\frac{( x + 4 ) 3}{4} = 6

15 - 0.5 q = 5

5^{\frac{1}{2}} = 5

What is the answer to t(n)=4t(n^2)+n^2log_{10}(n) ?
The answer to t(n)=4t(n^2)+n^2log_{10}(n) is t=(nlog_{10}(n))/(1-4n);\quad n\ne 0,n\ne 1/4