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.
Given:
The system of equations is
To find:
The correct statement for the given system of equations.
Solution:
On comparing the given equations and general form of linear equation, i.e., 
, we get
Here,
Since, 
, therefore, the two equations are equivalent and the system has infinitely many solutions.
Hence, the correct option is b.
Answer:
uh thats not really an actual question but k...
Step-by-step explanation:
there are no operations
Answer:
X>50
X>-51
Step-by-step explanation:
6+x>56
56+x>5
separate the equation into different equations by the sign
I) 6+x>56
x>56-6
x>50
ii) 56+x>5
x>5-56
x>-51
Answer:
186.3
Step-by-step explanation:
