Problem: We need to find the value of the logarithm log_(49)7. The base of the logarithm is 49, and the number we are taking the log of is 7. We are looking for the exponent that we need to raise 49 to in order to get 7.Definition of Logarithm: Recall the definition of a logarithm: if a^x = b, then log_a(b) = x. We need to find an exponent x such that 49^x = 7.Rewriting the Base: We know that 49 is a perfect square, specifically 49 = 7^2. Therefore, we can rewrite 49^x as (7^2)^x.Simplifying the Exponent: Using the property of exponents that (a^b)^c = a^(b*c), we can simplify (7^2)^x to 7^(2*x).Setting up the Equation: We want 7^(2*x) to equal 7, which is the same as 7^1. Therefore, we need 2*x to equal 1.Solving for x: To find x, we divide both sides of the equation 2*x = 1 by 2, which gives us x = 1/2.Final Result: Therefore, log_(49)7 is equal to 1/2 because raising 49 (which is 7^2) to the power of 1/2 gives us 7.

Grade 8

Relationship between squares and square roots

Full solution

\log _{49} 7=

Problem: We need to find the value of the logarithm $\log_{49}7$ . The base of the logarithm is $49$ , and the number we are taking the log of is $7$ . We are looking for the exponent that we need to raise $49$ to in order to get $7$ .
Definition of Logarithm: Recall the definition of a logarithm: if $a^x = b$ , then $\log_a(b) = x$ . We need to find an exponent $x$ such that $49^x = 7$ .
Rewriting the Base: We know that $49$ is a perfect square, specifically $49 = 7^2$ . Therefore, we can rewrite $49^x$ as $(7^2)^x$ .
Simplifying the Exponent: Using the property of exponents that $(a^b)^c = a^{(b*c)}$ , we can simplify $(7^2)^x$ to $7^{(2*x)}$ .
Setting up the Equation: We want $7^{2x}$ to equal $7$ , which is the same as $7^1$ . Therefore, we need $2x$ to equal $1$ .
Solving for x: To find x, we divide both sides of the equation $2x = 1$ by $2$ , which gives us $x = \frac{1}{2}$ .
Final Result: Therefore, $\log_{49}7$ is equal to $\frac{1}{2}$ because raising $49$ (which is $7^2$ ) to the power of $\frac{1}{2}$ gives us $7$ .