How do you solve ${log}_{b} 121 = 2$ $log_b121=2$ ?

Question

How do you solve ${log}_{b} 121 = 2$ $log_b121=2$ ?

1 Answer

Impact of this question

1932 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-05T06:35:51

$\pm 11$ $+-11$ , see explanation.

Explanation:

if you have $a^{b} = c$ $a^b = c$ , logarithms go ${log}_{a} c = b$ $log_ac = b$ . That's just a thing that you need to know, I'm not quite sure how to explain that, but if we apply it to the logarithm in the problem...

${log}_{b} 121 = 2 \to b^{2} = 121$ $log_b121=2 -> b^2=121$ .

And from here you just take the square root of both sides and get $\pm 11$ $+-11$ . It's plus or minus because $11^{2}$ $11^2$ and ${(- 11)}^{2}$ $(-11)^2$ equal 121. (Negative multiplied by a negative equals a positive).

Science

Math

Humanities

... and beyond

How do you solve ${log}_{b} 121 = 2$ $log_b121=2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve logb121=2log_b121=2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve ${log}_{b} 121 = 2$ $log_b121=2$ ?