How can I solve for x for
3^2x +5^x-1=-6
x = 4.6078
there's no easy algebraic way to solve that. A numeric method or graphical solution would be best.
Clearly, 4.6078 is wrong, since those powers of 3 and 5 are positive, so there's no way the sum can be -6
blurting out an answer serves no purpose, especially when it is incorrect
I suspect a typo
either
3^(2x)+3^x-1=-6
or
5^(2x) + 5^x-1=-6
As Steve pointed out, any power of 3^?? or 5^?? is always positive, so it can never be negative.
To solve the given equation for x, we can start by simplifying it.
First, let's rewrite the equation with the help of exponent rules. The exponent rule we'll use is:
a^(b + c) = a^b * a^c
Applying this rule to our equation, we get:
(3^2) * 3^x + 5^(x-1) = -6
Simplifying further, we have:
9 * 3^x + 5^(x-1) = -6
Now, let's break down the equation and isolate terms with exponents:
1) Solve terms with 3:
9 * 3^x = 9^1 * 3^x = 3^2 * 3^x = 3^(2 + x)
2) Solve terms with 5:
5^(x-1) = 5^x * 5^(-1) = 5^x / 5
3) Substituting our results back into the equation:
3^(2 + x) + 5^x / 5 = -6
Since we have a fraction, let's multiply both sides of the equation by 5 to eliminate it:
5 * (3^(2 + x) + 5^x / 5) = 5 * -6
This simplifies to:
3^(2 + x) + 5^x = -30
Now, we need to get rid of the exponent on 3. To do this, we can rewrite it as a power of the base 3:
3^(2 + x) = 3^2 * 3^x = 9 * 3^x
Substituting back into the equation, we get:
9 * 3^x + 5^x = -30
To proceed further, let me clarify the problem as there might be a mistake in it. In the equation you provided, the left side consists of exponential terms which are always positive, while the right side is negative (-6). This implies that there might not be any real solutions for x in this equation. Please double-check the equation provided or provide the correct equation for further analysis.