Find the base of lg(Cube root of 7) = 2/3?

Call the base b and exponentiate each side of the equation base b (take b to the power of each side). This is the inverse of the log base b, so they cancel out. This gives:

7^ 1/3 = b^ 2/3 = (b²)^ 1/3

b² = 7

b = ?7
 
log (base b) (cube roof of 7) = 2/3
log (base b) 7^(1/3) = 2/3
b^(2/3) = 7^(1/3)
(b^(2/3))^(3/2) = (7^(1/3))^(3/2)
b = 7^(1/2)
b = ?7
 
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