Question

help with these two probems? plzzz? get the exact and calculator answer

1. log(3)(7x-4)=2
2. 2log(4)3x+log(4)5x=log(4)225

Answer 1

log(3)(7x-4)=2 can be better written (and be less confusing) if done in Equation (below):\[\log_{3} (7x-4)=2\]

Answer 2

Since the log and the exponential functions are inverses, \[3^{\log_{3}(7x-4) }\]

Answer 3

boils down to 7x-4.
Now look at the other side of the given equation. It's 2. Use this 2 as the exponent of 3. Note how we're using the same base (3) on both sides of the equation. What do you get?

Answer 4

I don't understand

Answer 5

Have you studied the basic properties of exponential and logarithmic functions?
Did you know that \[\log_{a} a^x = x?\]

Answer 6

that's because the log and the exp function are inverses of one another.
Before I can help you well, I need to know what you undrstand and what you don't understand about the problem given you here.

log(3)(7x-4)=2 should be written as \[\log_{3} (7x-4) = 2. \]

Answer 7

Use the property of inverse functions to eliminate the operator (command) \[\log_{3} \]

Answer 8

So again, please tell me as best you can what you do and what you don't understand. We'll go from there.

Answer 9

1.\[\begin{align}
\log_3(7x-4) &= 2\\
7x-4 &= 3^2\\
7x &= 3^2+4\\
x &= \frac{3^2+4}7\\
&=
\end{align}\]