No.
Log x may be written more explicitly as log10(x). That is, the logarithm of x to the base 10.
Assuming that In x is a misprint for ln x, this is loge(x) ie the logarithm of e to the base e.
log10(x) = loge(x)/loge(10)
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You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.
X=W*(A)^T Use logarithms. T=log(A)/log(X/W)
log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."