TI-84 Plus Question.

Question

I am trying to do log on my TI-84 Plus, and it always equals 0 for any problem I do.

How would I do the problem log3(x)=-2? (base of 3)

Another one I had also was 4^(4x)=16^(x+1).

Thanks!

Answer 1

A. Log₃ ( x ) = -2 that means x = 3^-2 or equivalently x = 1/9 4^(4x)=16^(x+1).

B. 4^4x = 16^x+1

4^4x = 4^{2x +2}

4x = 2x +2

x = 1

Answer 2

It sounds like you might have a problem with your calculator. Try charging it or replacing the batteries. I would also try updating the calculator to the newest software if it isn't already fully updated. If it keeps happening, I would just go ahead and restore the factory settings on the calculator.

now, solving the problems:

(1) log₃x = -2

we need to figure out how to isolate the x variable

we know that;

if y = b^x , then log_by = x



so, we can rewrite:

log₃x = -2 ---------> 3^-2 = x

3^-2 = x

1 / (3^2) = x

x = 1/9

(2) 4^(4x)=16^(x+1)

this one seems trickier, but notice that 16 is actually a square of 4, so we can rewrite 16 as 4²

so;

4^(4x)=16^(x+1)

4^(4x)=4^2(x+1)

now, we can isolate the exponent terms because the bases are the same

4x = 2(x+1)

4x = 2x + 2

2x = 2

x = 1