how to solve for exponents?

we cannot solve such an equation.its 100% impossible

be conscious that 343 = 7^3 and 40 9 = 7^2 we are able to rewrite the difficulty as follows 7^(3(x^2 - 4)) = 7^(2*-3x) because of the fact the bases are equivalent we are able to now artwork in basic terms with the exponents (when you consider that 7^x is injective) 3(x^2 - 4) = 2*-3x 3x^2 - 12 = -6x 3x^2 + 6x - 12 = 0 x^2 + 2x - 4 = 0 x = (-2 +sqrt(4+sixteen))/2 or x = (-2-sqrt(4+sixteen))/2 x = (-2+2sqrt(5))/2 or x = (-2-2sqrt(5))/2 x = sqrt(5) - a million or x = -a million -sqrt(5) x = a million.23607 or x = -3.23607 be conscious that 36/35 = (6/5)^2 and a hundred twenty five/216 = (5^3/6^3)^-a million = (6/5)^-3 we rewrite the difficulty as (6/5)^(2(x+3)) = (6/5)^(-3(a million/(4-x))) we are able to equate exponents when you consider that bases are equivalent (and (6/5)^x is injective) we've 2(x+3) = -3/(4-x) 2(x+3)(4-x) = -3 -2x^2 + 2x + 12 = -3 2x^2 - 2x -15 = 0 x = (2+sqrt(4+a hundred and twenty))/4 or x = (2-sqrt(4 +a hundred and twenty))/4 x = 3.2839 or x = -2.2839

Firstly, that's not an equation so there's nothing to solve for. The only thing you can do is subtract the two 3^x terms.

3^x-3^-x-3^x

-3^-x

If you wanted something like 2^x = 4, then you can use logarithms.

log_2(4) = x

2 = x

In the example give 3^x cancels out leaving 3^-x

The usual way for solving exponent-related questions is to take Logs at both sides

e.g. a^b X c^d = E

gives b Log(a) + d Log(c) = Log(E)

the ans wud b simply -3^x

for finding the value of x we first need to equate it with some no. if its an eqn we can solve it by using log.

@ Mr. Brainy..... V Vivek

sir its -3^x not 3^(-x) so it can not be 1/3^x

3^x-3^-x-3^x is simply an expression. Unless t is set equal to something (i.e., in the form of an equation) there is nothing to solve. However, it can be simplified:

3^x-3^-x-3^x = 3^-x = 1/x^3

see, the answer is -3^x = 1/3^x

if d value of x is given then it's fine....otherwise generally logarithms are used in such problems.....

hope u understood......

3^x-3^-x-3^x

= 3^x and -3^x will cancel out

=> -3^-x