Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
First, let's find g(4). After we find it, we can substitute the value into 
.
To find g(4), we need to substitute x = 4 into the function 
, as shown below:
Now that we have found g(4) = 9, we can substitute this value in for 
in .
f(g(4)) is equal to 28.
Answer: X + X= 2x or X squared.
Step-by-step explanation:
Ooh so the directions tell us what we need to find the entire line, which is made up of FH. so just add GH (15) and FG (6) together to find FH. so that is 21. FH=21.
-7x+3y=-24+7x-2x+5y
-7x+3y=5x+5y-24
-12x-2y=-24
-7x+3y=-11
-36x-6y=-72
-14x+6y=-22
-50x=-94
Okay well I did something wrong
