Logarithmic problem: log_49(1/7) = y, 49^y = 1/7...?

[kt]

this is my question, use the algebraic definition of the logarithm to evaluate (simplify) the expression by hand and state the property numbers used.

〖log〗_49 (1/7)=y,
49〗^y=(1/7)

Is there more to this problem were.

 

[pka]

\(\displaystyle y=-\frac{1}{2}\)

 

[kt]

Can you explain, how you go to there please?

 

[pka]

\(\displaystyle \L 49^{\(\frac {-1}{2}\)}=\frac{1}{7}\)

 

[Mrspi]

Or a slightly different explanation. You had this:

49<SUP>y</SUP> = 1/7

We know that 1/7 can be written as 7<SUP>-1</SUP> by using the definition of a negative exponent. Now we have

49<SUP>y</SUP> = 7<SUP>-1</SUP>

Can we write 49 as a power of 7? Sure! 49 = 7<SUP>2</SUP>. So,

(7<SUP>2</SUP>)<SUP>y</SUP> = 7<SUP>-1</SUP>

When you raise a power to a power, you multiply the exponents. So,

7<SUP>2y</SUP> = 7<SUP>-1</SUP>

Since the bases are the same, and the expressions are equal, the exponents must be equal as well:

2y = -1

y = -1/2

 

[kt]

Thank you.