Solving exponential equations with logarithms?

Question

1.x= log^7(30)
2.f^(x^2-3) = 71

How do you do them right? Heres what i did, and i was wrong
log7+log30
.845 + 1.47= 2.3222
and for the second i did
log 72 divided by log 5 = xsquared - 3
which was 2.6572 = x squared - 3
then add 3 to get 5.6571 = x squared
then do the square root of 5.65.. which is 2.3784
but these are wrong can anyone tell me why?

Q2 is
5^(x^2-3) = 72

oh nvm 2 was right

Answer 1

I don't understand the way you've typed the question.

The symbol ^ means "exponent", which doesn't make sense here except where you put x^2.

A guess: Does number 1 mean :"log of 30 with base 7"?

If so, we can write it here as log(7) 30 or log(base 7) 30
or log 30 to base 7.
If it's that, calculate it as (log 30)/()log 7)
=1.747869697

While using Excel for that calculation just now, I notice they use
log(30,7) to mean log of 30 with base 7. Which makes me think that maybe q.1 is log of 7 in base 30, which is the reciprocal of the answer I've found.

Q.2 What you have done would give a correct solution to the equation
5^(x^2-3) = 72

I didn't work through to check your figures, but the process is correct and the numbers look reasonable. Is that the equation you had to solve? I notice there's a 71in the equation you typed for us, but I don't think the answer would be very different whether it was 71 or 72. So tell us, what was the equation really?

Answer 2

It looks like it should be (log^7)(30), so multiply by 30 instead of adding the log of both numbers together. Or is the base supposed to be 30, then it should be something like log30/log7 or vice versa, i don't fully remember.

Answer 3

1.x= log^7(30)

Assuming you mean log(base 7)(30) and not log(7 * 30), then

x = log(7)(30)
x = (log(30))/(log(7))
x = about 1.7479

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2.)
5^(x^2 - 3) = 72
(x^2 - 3)ln(5) = ln(72)
(ln(5))x^2 - 3ln(5) = ln(72)
(ln(5))x^2 = 3ln(5) + ln(72)
x^2 = (3ln(5) + ln(72))/(ln(5))

x = sqrt((3ln(5) + ln(72))/(ln(5))) or about ±2.3785