How do you evaluate #log_49 7#?

4 Answers
EZ as pi
Jul 27, 2016

#log_49 7 = 1/2#

Explanation:

Written in this form, the question being asked is.."How can I get 7 from 49?"

The answer is, by finding the square root. Another way of saying "square root" is the power of #1/2#.

In maths: #log_49 = x " "rArr 49^x = 7#

#49^(1/2) = 7#

Therefore #log_49 7 = 1/2#

We can also ask "Which power of 49 is equal to 7?:

Cesareo R.
Jul 27, 2016

#1/2#

Explanation:

#log_a b = (log_e b)/log_e a#

then

#log_{49}7 = (log_e 7)/(log_e 7^2)=(log_e7)/(2log_e 7)=1/2#

Tony B
Jul 27, 2016

Have a look at https://socratic.org/s/awwES2YN

#x=0.5#

Explanation:

#color(blue)("Derived by calculation")#

Set #log_49(7)=x#

Write as #49^x=7#.........................Equation(1)

Take logs to base 10 of each side of Equation(1)

#log_10(49^x)=log_10(7)#

This is the same as:

#xlog_10(49)=log_10(7)#

#x=log_10(7)/log_10(49)#

#x=0.5#

mason m
Jul 27, 2016

#1/2#

Explanation:

Let:

#x=log_49 7#

By the definition of logarithms, this can be rewritten as:

#49^x=7#

Write #49# as #7^2#:

#(7^2)^x=7#

Simplify #(7^2)^x# using the rule #(a^b)^b=a^(bc)#:

#7^(2x)=7#

Recall that #7=7^1#:

#7^(2x)=7^1#

Since the powers are equal, we know their bases also must be equal:

#2x=1#

#x=1/2#

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