help in solving log7 (1/49) = x

isn't 7^-2 = 1/49?

if so, log7 (1/49)=-2

log7 (1/49) = x is equivalent to

7^x = 1/49 by the definition of logarithms, thus ...
7^x = 49^-1
7^x = (7^2)^-1
x = -2

To solve the equation log7(1/49) = x, we need to use the logarithmic property that states:

log(base a) (b) = c if and only if a^c = b

In this case, we have log7(1/49) = x. Applying the logarithmic property, we rewrite the equation as:

7^x = 1/49

To simplify further, we can express 1/49 as a power of 7:

1/49 = 7^(-2)

Now, we have:

7^x = 7^(-2)

Since the bases (7) are the same on both sides of the equation, we can equate the exponents:

x = -2

Therefore, the solution to the equation log7(1/49) = x is x = -2.

To solve the equation log7 (1/49) = x, we need to use the definition of logarithms. Logarithms help us solve for an unknown exponent in an equation involving exponential functions.

In this case, the base of the logarithm is 7. The equation evaluates to:

log7 (1/49) = x

To get rid of the logarithm, we'll use the property of logarithms which states that for any base b, if log_b (x) = y, then b^y = x.

In this case, we can rewrite the equation as an exponential equation:

7^x = 1/49

Let's simplify 1/49. It is equivalent to 1 divided by 49, which gives us:

7^x = 1 ÷ 49

Using the rules of exponents, we know that 1 divided by any number is simply the number raised to the power of -1. So we rewrite the equation as:

7^x = 49^(-1)

To simplify further, we can rewrite 49^(-1) as 1/49:

7^x = 1/49

Now, both sides are fractions. We can write 1 as 7^0, as any number raised to the power of zero is equal to 1:

7^x = 7^0/49

Since the bases are the same, we can equate the exponents:

x = 0/49

Simplifying further, we have:

x = 0

Therefore, the solution to the equation log7 (1/49) = x is x = 0.