10 ^(2x + 1) = 4 ^ (x-1)?

10^(2x + 1) = 4^(x - 1)
log[10^(2x + 1)] = log[4^(x - 1)]
(2x + 1)log(10) = (x - 1)log(4)
2x + 1 = (x)log(4) - log(4)
2x - (x)log(4) = -1 - log(4)
x[2 - log(4)] = -1 - log(4)
x = [-1 - log(4)]/[2 - log(4)]
x = [log(4) + 1]/[log(4) - 2]
x = [log(4) + log(10)]/[log(4) - log(100)]
x = [log(40)]/[log(0.04)]

10 ^ (2x + 1) = 4 ^ (x - 1)

First using exponent laws, we can take out 1 and -1 from the respective exponents. Because

a^(b + c) = a^b * a^c, we can get

10 * 10 ^ 2x = (4^x) / 4

We can also take out the 2 from 2x and let 10^2 = 100

10 * 100 ^ x= (4 ^ x) / 4

Multiply both sides by 4

40 * 100 ^ x = 4 ^ x

Now also using the law that

a^c / b^c = (a/b)^c

We can divide both sides by 4^x, giving

40 * 25^x = 1

Divide both sides by 40

25^x = 1/40

Let 25^x = 5^2x

5^2x = 1/40

Take the natural logarithm of both sides:

ln(5 ^ 2x) = ln(1/40)

Let ln(1/40) = -ln(40) and ln(5 ^ 2x) = 2x(ln 5)

2x(ln 5) = - ln 40

2x = -ln 40 / ln 5

x = -ln 40 / (2ln 5)

Using a calculator:

x = -1.146

Checking the answer we get

10 ^ (-1.29) = 4^ (-2.14601)

0.0512 = 0.0512

So the answer is correct.

10^(2x + 1) = 4^(x - 1)
2x + 1 = log[4^(x - 1)]
2x + 1 = (x - 1)log(4)
xlog(4) - 2x = 1 + log(4)
x = [1 + log(4)] / [log(4) - 2]
x = log(40) / log(0.04)
x ≈ -1.146

( 2x + 1 ) log 10 = ( x - 1 ) log 4

Let log be log to base 10

2x + 1 = x log 4 - log 4

( 2 - log 4 ) x = - ( 1 + log 4 )

x = - ( 1 + log 4 ) / ( 2 - log 4 )

x = - 1.146