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3 years ago

If 5^3b-1=5^b-3 what is the value of b

Mathematics

2 answers:

Neporo4naja [7]3 years ago

8 0

Answer:

b=-1

Step-by-step explanation:

$5^{3b-1} = 5^{b-3}$

For any exponent, a^x

'a' is the base and x is the exponent

when the base are same on the both sides of equation then we equate the exponent

$3b-1=b-3$

we solve for b

Subtract 'b' on both sides

$2b -1= -3$

Add 1 on both sides

$2b = -2$

Divide by 2 on both sides

b=-1

andrezito [222]3 years ago

3 0

The answer is b=-1 hopefully the s up so you can see my work and understand how I did it.

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tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

Learn more about Solving Exponents at:

#learnwithBrainly

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