>>6675151) 5^(2x) -2*5^x-3 = 0

5^(2x) -2*5^x-3 = 0

(5^x)^2 -2*5^x-3 = 0

using u = 5^x, we have:

u^2-2u-3 = 0

Using the quadratic formula, we solve for u and get:

u1 = -1

u2 = 3

Substituting back u = 5^x, we have:

x1 : 5^x = -1

x2 : 5^x = 3

Using the log/exponent relationship log_b (a) = c <=> b^c = a, we have:

5^x = -1 => x = log_5 (-1) = no real solution

5^x = 3 => x = log_5 (3) = 0.68260619...

We can now verify the answer by plugging in the value in the original expression:

5^(2*0.68260619)-2*(5^0.68260619...)-3 = 0

0 = 0

2) log_4(3x+10) = log_4(x) + 1

using log_4(4) =1, we have:

log_4(3x+10) = log_4(x) + log_4(4)

Using the logarithmic product rule, we have:

log_4(x)+log_4(4) = log_4(x*4) = log_4(4x)

We end up with:

log_4(3x+10) = log_4(4x)

log_4(3x+10) -log_4(4x) = 0

Using the quotient rule, we have:

log_4((3x+10)/4x) = 0

We get rid of the log using the log/exponent relationship log_b (a) = c <=> b^c = a:

log_4((3x+10)/4x) = 0 => 4^0 = (3x+10)/4x

1 = (3x+10)/4x

4x = 3x+10

4x-3x = 10

x = 10

We can now verify x=10 by plugging it in the original expression:

log_4(3(10)+10) = log_4(10) +1

log_4(40) = log_4(10)+1

2.66.... = 2.66...

Hope this helps.