But That's Not What I Got!

Algebra Level 4

log ⁡ 4 ( x − 1 ) = log ⁡ 2 ( x − 3 ) \large\log_4(x-1)=\log_2(x-3) Find the sum of all possible values of x x satisfying the equation above.

5 6 None of these. 7 4 8 9

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
May 28, 2016

Relevant wiki: Logarithms

log ⁡ 4 ( x − 1 ) = log ⁡ 2 ( x − 3 ) 1 2 log ⁡ 2 ( x − 1 ) = log ⁡ 2 ( x − 3 ) log ⁡ 2 ( x − 1 ) = log ⁡ 2 ( x − 3 ) 2 ( x − 1 ) = ( x − 3 ) 2 x 2 − 7 x + 10 = 0 ( x − 5 ) ( x − 2 ) = 0 \begin{aligned} \log_4(x-1)&=\log_2(x-3)\\ \frac 12\log_2(x-1)&=\log_2(x-3)\\ \log_2(x-1)&=\log_2(x-3)^2\\ (x-1)&=(x-3)^2\\ x^2-7x+10&=0\\ (x-5)(x-2)&=0 \end{aligned}

Therefore, x = 5 x=5 or x = 2 x=2 . But, observe that when x = 2 x=2 , the RHS of the equation is not defined, since the logarithm function is defined on positive reals only. Thus, the equation has only one solution that is 5 5 .

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Moderator note:

Thanks for checking that the solutions to the manipulated equation actually satisfy the original equation.

@Sravanth Chebrolu , I have been editing your problems. Please don't key in semicolon (;) instead of colon (:). Even colon is usually unnecessary. Thanks.

Chew-Seong Cheong - 5 years ago

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Oh sorry sir, I'll take care of that.

Sravanth C. - 5 years ago

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