5 answers
To solve the equation \( x \cdot \frac{1}{5} = 6 \), you need to isolate \( x \).
You can do this by multiplying both sides of the equation by 5:
\[ x \cdot \frac{1}{5} \cdot 5 = 6 \cdot 5 \]
This simplifies to:
\[ x = 30 \]
Therefore, the missing value is \( x = 30 \).
To solve the equation \( 2^{(x+2)} = 4^2 \), we first rewrite \( 4^2 \) in terms of base 2:
\[ 4 = 2^2 \implies 4^2 = (2^2)^2 = 2^{2 \cdot 2} = 2^4 \]
Now, we can set the exponents equal to each other since the bases are the same:
\[ x + 2 = 4 \]
Next, solve for \( x \) by subtracting 2 from both sides:
\[ x = 4 - 2 = 2 \]
Thus, the missing value is \( x = 2 \).
To solve the equation \( 3^{(x-2)} = 81 \), we first express \( 81 \) as a power of \( 3 \):
\[ 81 = 3^4 \]
Now the equation becomes:
\[ 3^{(x-2)} = 3^4 \]
Since the bases are the same, we can set the exponents equal to each other:
\[ x - 2 = 4 \]
Next, solve for \( x \) by adding 2 to both sides:
\[ x = 4 + 2 = 6 \]
Thus, the missing value is \( x = 6 \).
