Prove that log₁₀₀(x-1)=log₁₀(x-3)
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I assume you are familiar with the property that you can shift the power of base in multiplication of log but in reciprocal as follows
logbase10^2 (x-1) = logbase 10 (x-3)
1/2logbase10 (x-1) = logbase 10 (x-3)
now we know n*logm = log m^n
so logbase10 (x-1) = logbase 10 (x-3)^2
so (x-1) = (x-3)^2
(x-1) = x^2 -6x +9
so x^2-7x +10 =0
by solving above quadratic equation we get
x=2,5 answer
please mark as brainlist
logbase10^2 (x-1) = logbase 10 (x-3)
1/2logbase10 (x-1) = logbase 10 (x-3)
now we know n*logm = log m^n
so logbase10 (x-1) = logbase 10 (x-3)^2
so (x-1) = (x-3)^2
(x-1) = x^2 -6x +9
so x^2-7x +10 =0
by solving above quadratic equation we get
x=2,5 answer
please mark as brainlist
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