log(x + 1) = 2, x =? Understand Why Logarithms are so POWERFUL!

[@stevendebettencourt7651]

One thing I want to make clear before I start solving this is that when we see “log(a)” in an equation with no base number with it, I take it to be log base 10. If we are talking about any other base, that has to be made explicit in the equation. And of course, if we are taking the log base e, we use “ln(a)” for this, the natural log. With that out of the way, let’s get started. log(x + 1) = 2 Remember that logarithms are directly related to exponents via the following: log base a (b) = c a^c = b So we can restate the above equation: log(x + 1) = 2 10^2 = x + 1 And then it’s just doing the math and solving for x: => x + 1 = 100 => x = 99 So the answer is A.

[@Stylux-z1p]

When the base is not specified, log typically refers to the base-10 logarithm -> log_10 log(x + 1) = 2 lets try the exponential form: 10² = x + 1 100 = x + 1 99 = x --> a) ✔ -----------------------------------------------------

[@thomasharding1838]

You are WRONG!!! When the base is not specified "log" ABSOLUTELY refer to base 10!! (Yes, I know, I'm being a smart ass but I can't help it, all of me is smart, and I did give you a Thumbs Up!)

[@charlesmitchell5841]

You can assume that the base is 10 since it is not stated, so 10^2=100 and 99+1=100 so a)99 would be the correct answer I believe.

[@detlefkretschmer2917]

I agree with the other comments, log alone is wrong in the context. There is actually a standard for this, log to the base 10 is supposed to be written as lg, to the base x as log subscript x. But of course, Americans don't have to keep to standards.

[@wes9627]

Many Will Get WRONG! because the problem is poorly stated. Students, if the question isn't clear, state your assumptions in arriving at an answer.

[@armchairtin-kicker503]

log(x+1) = 2 Exponentiate both sides of the equation by a base of 10, canceling the log-function on the LHS... 10^(log(x+1)) = 10^2 x + 1 = 100 (x +1) -1 = (100) -1 x = 99 Therefore the answer is "a."

[@mehdizangiabadi-iw6tn]

(log10^2-log10*log10^.53)+log10^.53

[@russelllomando8460]

couldn't remember all the processes but by process of elimination....99. thanks for the fun

[@chrisdissanayake6979]

Answer: a) 99 -------- Iog(x+1) = 2 10^2 = x+1 x+1 = 100 x = 100-1 = 99 x = 99

[@papuan-black]

First person to like and comment 11th December 2024 ❤

[@tomtke7351]

Log_10 of 2 = x where 10^x = 2 log(x+1)=2 means 10^2=x+1 100=x+1 x=99 VERIFY log(x+1)=?2 log(99+1)=?2 log100=?2 2=❤2✔️

[@mehdizangiabadi-iw6tn]

9999=7641 ???? How does

[@pspprabhat]

10^2=×+1&×=99

[@josephlaura7387]

a) 99

[@adgf1x]

x=99

[@mauriziograndi1750]

X=99 because log100=2

[@daniellerosalie2155]

X=99