How do you solve $log (2 x + 1) = 6 + log (x - 1)$ $Log(2x+1)=6+log(x-1)$ ?

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How do you solve $log (2 x + 1) = 6 + log (x - 1)$ $Log(2x+1)=6+log(x-1)$ ?

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Answer 1 · 2016-02-12T03:54:32

Since the logs are in a common base we can put all logs to one side an simplify by using the logarithm law ${log}_{a} n - {log}_{a} m = {log}_{a} (\frac{n}{m})$ $log_an - log_am = log_a(n/m)$

Explanation:

$log (2 x + 1) - log (x - 1) = 6$ $log(2x + 1) - log(x - 1) = 6$

$log (\frac{2 x + 1}{x - 1}) = 6$ $log((2x + 1)/(x - 1)) = 6$

Since nothing is noted in subscript, the log is in base 10.

$10^{6} = (\frac{2 x + 1}{x - 1})$ $10^6 = ((2x + 1)/(x - 1))$

$1000000 (x - 1) = 2 x + 1$ $1 000 000(x - 1) = 2x + 1$

$1000000 x - 1000000 = 2 x + 1$ $1 000 000x - 1 000 000 = 2x + 1$

$999998 x = 1000001$ $999 998x = 1 000 001$

$x = (\frac{1000001}{999998})$ $x = (1000001/(999 998))$

Your teacher may want you to keep the answer in fractional form, but he/she may want it rounded off, so keep that in mind.

Practice exercises:

Solve for x. Leave answers in exact form.

a) ${log}_{7} x = 3$ $log_7x = 3$

b) ${log}_{3} (x + 1) = 5 - {log}_{3} (4 x^{2} - 4)$ $log_3(x + 1) =5 - log_3(4x^2 - 4)$

Good luck!

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How do you solve $log (2 x + 1) = 6 + log (x - 1)$ $Log(2x+1)=6+log(x-1)$ ?

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Explanation:

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