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solve for x (3 dp)
a) 3^x=19.
b) 2^1-x=7
c) (1/4)^x=75.
How to do this?

You need to use logarithms

a)
if 3 ^ x = 19
x = log 19 to the base 3
x = 2.68

for b)
2^1 = 2
2 - x =7
-x = 7-2
x = -5

for c)
if (1/4) ^ x = 75
x = log 75 to the base 0.25
x = -3.11

I think for b, I got 1.807.. and for a and c I got the same answer as yours.

the formula is
if x^y = z
y = log z to the base x

hm.. ok.

You understand logarithms right

hm.. yup I understand it.

So then you should show this as your method

1/2 ?

nope. she has the advantage of going first, so it must be greater than 1/2.

No
i think it should just be 0.75
I can't think of anything else

Whats your answer and method