Find the value of n where (225)base n = 149

Question

Find a value of x where 4^x and 5^x consist of the same digits in different order

Answer 1

8

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225 (base n)=(2*n*n+2*n+5)
2n^2+2n+5=149
solving this we get n=8

Answer 2

225 in base n = 2×n^2 + 2×n + 5 = 149 in base 10
2n^2 + 2n + 5 = 149
2n^2 + 2n - 144 = 0
n^2 + n - 72 = 0
n^2 + 9n - 8n - 72 = 0
n×(n + 9) - 8(n + 9) = 0
n - 8 = 0
n = 8.