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Sagot :
To solve the equation [tex]\( \log_{49} x = \frac{1}{2} \)[/tex], we need to rewrite it in an exponential form. Recall that the logarithmic statement [tex]\( \log_b (a) = c \)[/tex] can be rewritten as the exponential statement [tex]\( b^c = a \)[/tex].
Given the equation:
[tex]\[ \log_{49} x = \frac{1}{2} \][/tex]
we can rewrite it as:
[tex]\[ 49^{\frac{1}{2}} = x \][/tex]
Next, we need to simplify [tex]\( 49^{\frac{1}{2}} \)[/tex]. The expression [tex]\( 49^{\frac{1}{2}} \)[/tex] is equivalent to the square root of 49:
[tex]\[ \sqrt{49} = x \][/tex]
Since the square root of 49 is 7, we can simplify this as:
[tex]\[ x = 7 \][/tex]
Thus, the solution to the equation [tex]\( \log_{49} x = \frac{1}{2} \)[/tex] is:
[tex]\[ x = 7 \][/tex]
Given the equation:
[tex]\[ \log_{49} x = \frac{1}{2} \][/tex]
we can rewrite it as:
[tex]\[ 49^{\frac{1}{2}} = x \][/tex]
Next, we need to simplify [tex]\( 49^{\frac{1}{2}} \)[/tex]. The expression [tex]\( 49^{\frac{1}{2}} \)[/tex] is equivalent to the square root of 49:
[tex]\[ \sqrt{49} = x \][/tex]
Since the square root of 49 is 7, we can simplify this as:
[tex]\[ x = 7 \][/tex]
Thus, the solution to the equation [tex]\( \log_{49} x = \frac{1}{2} \)[/tex] is:
[tex]\[ x = 7 \][/tex]
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