Solving this Logarithmic Function?

Note that:

3^(2x + 1) - 7(3^x) + 4 = 0

==> [3^(2x) * 3^1] - 7(3^x) + 4 = 0

==> 3[3^(2x)] - 7(3^x) + 4 = 0.

Letting u = 3^x yields:

3u^2 - 7u + 4 = 0

==> (3u - 4)(u - 1) = 0

==> u = 4/3 and u = 1.

Then since u = 3^x, we see that:

3^x = 4/3 ==> x = log₃(4/3)/3 ≈ 0.087

3^x = 1 ==> 3^x = 3^0 ==> x = 0.

I hope this helps!

when x = 0 you have 3¹ - 7•3º + 4 = 3 - 7 + 4 = 0

graphing suggests no other solutions

the common logarithmic function has a base of 10. As you extra your study you will see that different logarithmic applications could have a base such by way of fact the extensive type 2 and could be written as log(base small 2)x. The inverse of the log is exponentiation. hence, if the backside is ten you will possibly have 10^(y/3). additionally be conscious that by way of fact the backside differences so will the exponential. So if it have been 2, it may be 2^(x-a million) or despite.

There are 2 roots:

x = 0

and

x = log(4) / log(3) - 1

(solve the equation as if it was a second degree equation where the variable is 3^x)