Question

(x-1)^lg2x-lgx^2=(x-1)^3

Answer 1

I hope this helps you

Answer 2

To solve the equation $(x-1)^{\log_2x} - \log_x^2 = (x-1)^3$, we'll first try to simplify it step by step.

1. Express $\log_2x$ as $\frac{\log x}{\log 2}$: $(x-1)^{\frac{\log x}{\log 2}} - \log_x^2 = (x-1)^3$

2. Get rid of the logarithm on the left side by raising both sides to the power of $\log 2$: $\left((x-1)^{\frac{\log x}{\log 2}}\right)^{\log 2} = \left((x-1)^3\right)^{\log 2}$

3. Simplify both sides: $(x-1)^{\log x} = (x-1)^{3\log 2}$

Now, we have a common base $(x-1)$ raised to different exponents. For two numbers to be equal when raised to the same exponent, the exponents must be equal. So, we can set the exponents equal to each other:

$\log x = 3\log 2$

Now, let's solve for $x$:

$\log x = \log 2^3$

Using the property $\log_b(a^c) = c\log_b(a)$:

$\log x = \log 8$

Now, since the logarithm function is one-to-one, we can remove the logarithms from both sides:

$x = 8$

Thus, the solution to the equation is $x = 8$.