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Olve for x log base 7 of x+ log base 7 of 6x-1=1

User RPinel
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Final answer:

To solve the equation log base 7 of x + log base 7 of 6x-1 = 1, combine the logarithms and simplify the expression. Rewrite the equation in exponential form and solve as a quadratic equation. The solutions are x = 1 and x = -7/6.

Step-by-step explanation:

To solve for x, we can use logarithmic properties. First, combine the two logarithms using the property log(a) + log(b) = log(ab). This gives us log7(x(6x-1)) = 1.
Next, simplify the expression inside the logarithm by multiplying x and (6x-1), resulting in log7(6x2-x) = 1.
Now, we can rewrite the equation using the exponential form of logarithms. Since the base is 7, the equation becomes 71 = 6x2-x.
Simplifying further, we get 7 = 6x2-x.
This is a quadratic equation that can be solved by setting it equal to zero: 6x2-x-7 = 0.
Using factoring or the quadratic formula, we find the solutions x = 1 and x = -7/6. Therefore, the values of x that satisfy the equation log7(x) + log7(6x-1) = 1 are x = 1 and x = -7/6.

User Sucharu Hasija
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