Question 1119200
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log[base2] of a number x is twice log[base4] of the same number x, because 4 is 2 squared:<br>
log[base 4}x = n  -->   x = 4^n = (2^2)^n = 2^(2n)  -->  log[base2]x = 2n<br>
Your book says to change log[base4]x to an equivalent expression involving log[base2].  Since log[base4] of a number is half of log[base2] of the same number, that would introduce fractions into the calculations.<br>
I would much rather change the log[base2] expression into an equivalent expression using log[base4].<br>
{{{log(2,(x+1))-log(4,x) = 1}}}    given equation
{{{2*log(4,(x+1))-log(4,x) = 1}}}  convert log base 2 to log base 4
{{{log(4,(x+1)^2)-log(4,x) = 1}}}  n*log(x) = log(x^n)
{{{log(4,(((x+1)^2)/x)) = 1}}}     log(a)-log(b) = log(a/b)
{{{((x+1)^2)/x = 4^1 = 4}}}        definition of logarithms
{{{x^2+2x+1 = 4x}}}
{{{x^2-2x+1 = 0}}}
{{{(x-1)^2 = 0}}}
{{{x = 1}}}